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Question 1:

Which of the following are quadratic equation in x ?
(i) x2-x+3=0
(ii) 2x2+52x-3=0
(iii) 2x2+7x+52=0
(iv) 13x2+15x-2=0
(v) x2-3x-x+4=0
(vi) x-6x=3
(vii) x+2x=x2
(viii) x2-1x2=5
(ix) x+23=x3-8
(x) 2x+33x+2=6x-1x-2
(xi) x+1x2=2x+1x+3

Answer:

i) (x2  x + 3) is a quadratic polynomial. x2  x + 3 = 0 is a quadratic equation.ii) Clearly, (2x2 + 52x  3) is a quadratic polynomial. 2x2 + 52x  3 = 0 is a quadratic equation.iii) Clearly, (2x2 + 7x + 52) is a quadratic polynomial. 2x2 + 7x + 52 = 0 is a quadratic equation.iv) Clearly, (13x2 + 15x  2) is a quadratic polynomial. 13x2 + 15x  2 = 0 is a quadratic equation.v) (x2  3x  x + 4) contains a term with x, i.e, x12, where 12 is not a integer. Therefore, it is not a quadratic polynomial. x2  3x  x + 4 = 0 is not a quadratic equation.vi) x  6x = 3 x2  6 = 3x x2  3x  6 = 0(x23x6) is a quadratic polynomial; therefore, the givenequation is quadratic. vii) x + 2x = x2 x2 + 2 = x3 x3  x2  2 = 0(x3  x2  2) is  not a quadratic polynomial. x3  x2  2 = 0 is not a quadratic equation.viii) x2  1x2 = 5 x4  1 = 5x2 x4  5x2  1 = 0(x4  5x2  1) is a polynomial with degree 4. x4  5x2  1 = 0 is not a quadratic equation.
(ix) x+23=x3-8
x3+6x2+12x+8=x3-86x2+12x+16=0
This is of the form ax2 + bx + c = 0.
Hence, the given equation is a quadratic equation.
(x) 2x+33x+2=6x-1x-2
6x2+4x+9x+6=6x2-3x+26x2+13x+6=6x2-18x+1231x-6=0
This is not of the form ax2 + bx + c = 0.
Hence, the given equation is not a quadratic equation.
(xi) x+1x2=2x+1x+3
x2+1x2=2x2+1x+3x2+12=2xx2+1+3x2x4+2x2+1=2x3+2x+3x2x4-2x3-x2-2x+1=0
This is not of the form ax2 + bx + c = 0.
Hence, the given equation is not a quadratic equation.

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Question 2:

Which of the following are the roots of 3x2+2x-1=0 ?
(i) −1
(ii) 13
(iii) -12

Answer:

The given equation is (3x2 + 2x  1 = 0).(i) x = (1) L.H.S. = x2 + 2x  1= 3 × (1)2 + 2 × (1)  1= 3  2  1= 0= R.H.S.Thus, (1) is a root of (3x2 + 2x  1 = 0).(ii) On substituting x = 13 in the given equation, we get:L.H.S. = 3x2 + 2x  1  = 3 × 132 + 2 × 13  1 = 3 1× 193+ 23  1= 1 + 2  33 = 03= 0= R.H.S.Thus,  13 is a root of (3x2 + 2x  1 = 0).(iii) On substituting x = 12 in the given equation, we get:L.H.S. = 3x2 + 2x  1  = 3 × 122 + 21 × 121  1 = 3 × 14 1 1= 34  2= 3  84= 54  0Thus, L.H.S.= R.H.S.Hence, 12 is a not solution of  (3x2 + 2x  1= 0).

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Question 3:

Find the value of k for which x = 1 is root of the equation x2+kx+3=0.

Answer:

It is given that (x=1) is a root of (x2 + kx + 3 = 0). Therefore, (x=1) must satisfy the equation. (1)2 + k × 1 + 3 = 0  k + 4 = 0  k = 4Hence, the required value of k is 4.

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Question 4:

Find the values of a and b for which x=34 and x=-2 are the roots of the equation ax2+bx-6=0.

Answer:

It is given that 34 is a root of ax2 + bx  6 = 0; therefore, we have:a × (34)2 + b × 34 6 = 0 9a16 + 3b4 = 6 9a + 12b16 = 6 9a + 12b  96 = 0 3a + 4b = 32        ...(i) Again, (2) is a root of ax2 + bx  6 = 0; therefore, we have:a×(2)2 + b×(2)  6 = 0 4a  2b = 6 2a  b = 3        ...(ii)On multiplying (ii) by 4 and adding the result with (i), we get: 3a + 4b + 8a  4b = 32 + 12 11a = 44 a = 4Putting the value of a in  (ii), we get:2×4  b = 3  b = 3 b = 5Hence, the required values of a and b are 4 and 5, respectively.

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Question 5:

Solve each of the following quadratic equations:

(2x − 3)(3x + 1) = 0

Answer:

(2x − 3)(3x + 1) = 0

⇒ 2x − 3 = 0 or 3x + 1 = 0

⇒ 2x = 3 or 3x = −1

x32 or x = -13

Hence, the roots of the given equation are 32 and -13.

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Question 6:

Solve each of the following quadratic equations:

4x2 + 5x = 0

Answer:

4x2 + 5x = 0

x(4x + 5) = 0

x = 0 or 4x + 5 = 0

x = 0 or x = -54

Hence, the roots of the given equation are 0 and -54.

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Question 7:

3x2-243=0

Answer:

Given:3x2  243 = 0 3(x2  81) = 0 (x)2  (9)2 = 0 (x + 9)(x  9) = 0 x + 9 = 0 or x  9 = 0 x = 9 or  x= 9Hence,9 and 9 are the roots of the equation 3x2243=0.

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Question 8:

Solve each of the following quadratic equations:

2x2+x-6=0

Answer:

We write, x=4x-3x as 2x2×-6=-12x2=4x×-3x

 2x2+x-6=02x2+4x-3x-6=02xx+2-3x+2=0x+22x-3=0

x+2=0 or 2x-3=0x=-2 or x=32
Hence, the roots of the given equation are -2 and 32.

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Question 9:

Solve each of the following quadratic equations:

x2+6x+5=0

Answer:

We write, 6x=x+5x as x2×5=5x2=x×5x

 x2+6x+5=0x2+x+5x+5=0xx+1+5x+1=0x+1x+5=0

x+1=0 or x+5=0x=-1 or x=-5

Hence, the roots of the given equation are −1 and −5.

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Question 10:

Solve each of the following quadratic equations:

9x2-3x-2=0

Answer:

We write, -3x=3x-6x as 9x2×-2=-18x2=3x×-6x

 9x2-3x-2=09x2+3x-6x-2=03x3x+1-23x+1=03x+13x-2=0
3x+1=0 or 3x-2=0x=-13 or x=23
Hence, the roots of the given equation are -13 and 23.

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Question 11:

x2+12x+35=0

Answer:

Given:x2 + 12x + 35 = 0 x2 + 7x + 5x + 35 = 0 x(x + 7) + 5(x + 7) = 0 (x + 5)(x + 7) = 0 x + 5 = 0 or x + 7 = 0 x = 5 or x = 7Hence,5 and 7 are the roots of the equation x2 + 12x + 35 = 0.

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Question 12:

x2=18x-77

Answer:

Given:x2 = 18x  77 x2  18x + 77 = 0  x2  (11x + 7x) + 77 = 0 x2  11x  7x + 77 = 0 x(x  11)  7(x  11) = 0 (x  7)(x  11) = 0 x  7 = 0 or x  11 = 0 x = 7 or x = 11Hence, 7 and 11 are the roots of the equation x2 = 18x  77.

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Question 13:

6x2+11x+3=0

Answer:

Given:6x2 + 11x + 3 = 0 6x2 + 9x + 2x + 3 = 0 3x(2x + 3) + 1(2x + 3) = 0 (3x + 1)(2x + 3) = 0 3x + 1 = 0 or 2x + 3 = 0 x = 13 or x = 32Hence, 13 and 32are the roots of the equation 6x2 + 11x + 3 = 0.

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Question 14:

6x2+x-12=0

Answer:

Given:6x2 + x  12 = 0 6x2 + 9x  8x  12 = 0 3x(2x + 3)  4(2x + 3) = 0 (3x  4)(2x + 3) = 0 3x  4 = 0 or 2x + 3 = 0 x = 43 or x = 32Hence, 43 and 32 are the roots of the equation 6x2 + x  12 = 0.

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Question 15:

Solve each of the following quadratic equations:

3x2-2x-1=0

Answer:

We write, -2x=-3x+x as 3x2×-1=-3x2=-3x×x

 3x2-2x-1=03x2-3x+x-1=03xx-1+1x-1=0x-13x+1=0
x-1=0 or 3x+1=0x=1 or x=-13
Hence, the roots of the given equation are 1 and -13.

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Question 16:

4x2-9x=100

Answer:

Given:4x2  9x = 100 4x2  9x  100 = 0 4x2  (25x  16x)  100 = 0 4x2  25x + 16x  100 = 0 x(4x  25) + 4(4x  25) = 0 (4x  25)(x + 4) = 0 4x  25 = 0 or x + 4 = 0 x = 254 or x = 4Hence, the roots of the equation are 254 and 4.

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Question 17:

15x2-28=x

Answer:

Given:15x2  28 = x 15x2  x  28 = 0 15x2 (21x  20x)  28 = 0 15x2  21x + 20x  28 = 0 3x(5x  7) + 4(5x  7) = 0 (3x + 4)(5x  7) = 0 3x + 4 = 0 or 5x  7 = 0 x = 43 or x = 75Hence, the roots of the equation are 43 and 75.

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Question 18:

4-11x=3x2

Answer:

Given: 11x = 3x2  3x2 + 11x  4 = 0 3x2 + 12x  x  4 = 0 3x(x + 4)  1(x + 4) = 0 (x + 4)(3x  1) = 0 x + 4 = 0 or 3x  1 = 0 x = 4 or x = 13Hence, the roots of the equation are 4 and 13.



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Question 19:

48x2-13x-1=0

Answer:

Given:48x2  13x  1 = 0 48x2  (16x  3x)  1 = 0 48x2  16x + 3x  1 = 0 16x(3x  1) + 1(3x  1) = 0 (16x + 1)(3x  1) = 0 16x + 1 = 0 or 3x  1 = 0 x = 116 or x = 13Hence, the roots of the equation are 116 and 13.

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Question 20:

Solve each of the following quadratic equations:

x2+22x-6=0

Answer:

We write, 22x=32x-2x as x2×-6=-6x2=32x×-2x

 x2+22x-6=0x2+32x-2x-6=0xx+32-2x+32=0x+32x-2=0
x+32=0 or x-2=0x=-32 or x=2
Hence, the roots of the given equation are -32 and 2.

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Question 21:

Solve each of the following quadratic equations:

3x2+10x+73=0

Answer:

We write, 10x=3x+7x as 3x2×73=21x2=3x×7x

3x2+10x+73=03x2+3x+7x+73=03xx+3+7x+3=0x+33x+7=0
x+3=0 or 3x+7=0x=-3 or x=-73=-733
Hence, the roots of the given equation are -3 and -733.

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Question 22:

3x2+11x+63=0

Answer:

Given:3x2 + 11x + 63 = 0 3x2 + 9x + 2x + 63 = 0 3x(x + 33) + 2(x + 33) = 0 (x + 33)(3x + 2) = 0 x + 33 = 0 or 3x + 2 = 0 x = 33 or x = 23 = 2 × 33 × 3 = 233Hence, the roots of the equation are 33 and 233.

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Question 23:

37x2+4x-7=0

Answer:

Given:37x2 + 4x  7 = 0 37x2 + 7x  3x  7 = 0 7x(3x + 7)  1(3x + 7) = 0 (3x + 7)(7x  1) = 0 3x + 7 = 0 or 7x  1 = 0 x = 73 or x = 17 = 1 × 77 × 7 = 77Hence, the roots of the equation are 73 and 77.

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Question 24:

Solve each of the following quadratic equations:

7x2-6x-137=0

Answer:

We write, -6x=7x-13x as 7x2×-137=-91x2=7x×-13x

7x2-6x-137=07x2+7x-13x-137=07xx+7-13x+7=0x+77x-13=0
x+7=0 or 7x-13=0x=-7 or x=137=1377
Hence, the roots of the given equation are -7 and 1377.

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Question 25:

46x2-13x-26=0

Answer:

Given:46x2  13x  26 = 0 46x2  16x + 3x  26 = 0 42x(3x  22) + 3(3x  22) = 0 (42x + 3)(3x  22) = 0 42x + 3 = 0 or 3x  22 = 0 x = 342 = 3 × 242 × 2 68 or x = 223 = 22 × 33 × 3 = 263Hence, the roots of the equation are 68 and 263.

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Question 26:

Solve each of the following quadratic equations:

3x2-26x+2=0                             [CBSE 2010, 12]

Answer:

We write, -26x=-6x-6x as 3x2×2=6x2=-6x×-6x

3x2-26x+2=03x2-6x-6x+2=03x3x-2-23x-2=03x-23x-2=0
3x-22=03x-2=0x=23 =63
Hence, 63 is the repreated root of the given equation.

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Question 27:

Solve each of the following quadratic equations:

3x2-22x-23=0                  [CBSE 2011]

Answer:

We write, -22x=-32x+2x as 3x2×-23=-6x2=-32x×2x

3x2-22x-23=03x2-32x+2x-23=03xx-6+2x-6=0x-63x+2=0
x-6=0 or 3x+2=0x=6 or x=-23 =-63
Hence, the roots of the given equation are 6 and -63.

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Question 28:

Solve each of the following quadratic equations:

x2-35x+10=0                  [CBSE 2011]

Answer:

We write, -35x=-25x-5x as x2×10=10x2=-25x×-5x

x2-35x+10=0x2-25x-5x+10=0xx-25-5x-25=0x-25x-5=0
x-5=0 or x-25=0x=5 or x=25
Hence, the roots of the given equation are 5 and 25.

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Question 29:

Solve each of the following quadratic equations:

x2-3+1x+3=0        [CBSE 2015]

Answer:

x2-3+1x+3=0x2-3x-x+3=0xx-3-1x-3=0x-3x-1=0
x-3=0 or x-1=0x=3 or x=1
Hence, 1 and 3 are the roots of the given equation.

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Question 30:

Solve each of the following quadratic equations:

x2+33x-30=0        [CBSE 2015]

Answer:

We write, 33x=53x-23x as x2×-30=-30x2=53x×-23x

x2+33x-30=0x2+53x-23x-30=0xx+53-23x+53=0x+53x-23=0
x+53=0 or x-23=0x=-53 or x=23
Hence, the roots of the given equation are -53 and 23.

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Question 31:

Solve each of the following quadratic equations:

2x2+7x+52=0       [CBSE 2013]

Answer:

We write, 7x=5x+2x as 2x2×52=10x2=5x×2x

2x2+7x+52=02x2+5x+2x+52=0x2x+5+22x+5=02x+5x+2=0
x+2=0 or 2x+5=0x=-2 or x=-52=-522
Hence, the roots of the given equation are -2 and -522.

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Question 32:

Solve each of the following quadratic equations:

5x2+13x+8=0                  [CBSE 2013C]

Answer:

We write, 13x = 5x + 8x as 5x2×8=40x2=5x×8x
 5x2+13x+8=05x2+5x+8x+8=05xx+1+8x+1=0x+15x+8=0
x+1=0 or 5x+8=0x=-1 or x=-85
Hence, -1 and -85 are the roots of the given equation.

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Question 33:

43x2+5x-23=0

Answer:

Given:x2  (1 + 2)x + 2 = 0 x2  x  2x + 2 = 0 x(x  1)  2(x  1) = 0 (x  2)(x  1) = 0 x  2 = 0 or x  1 = 0 x = 2 or x = 1Hence, the roots of the equation are 2 and 1.

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Question 34:

9x2+6x+1=0

Answer:

Given:9x2 + 6x + 1 = 0 9x2 + 3x + 3x + 1 = 0 3x(3x + 1) + 1(3x + 1) = 0 (3x + 1)(3x + 1) = 0 3x + 1 = 0 or 3x + 1 = 0 x = 13 or x = 13Hence, 13is the root of the equation 9x2 + 6x + 1 = 0.

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Question 35:

Solve each of the following quadratic equations:

100x2-20x+1=0

Answer:

We write, -20x=-10x-10x as 100x2×1=100x2=-10x×-10x
 100x2-20x+1=0100x2-10x-10x+1=010x10x-1-110x-1=010x-110x-1=0
10x-12=010x-1=0x=110
Hence, 110 is the repreated root of the given equation.

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Question 36:

Solve each of the following quadratic equations:

2x2-x+18=0

Answer:

We write, -x=-x2-x2 as 2x2×18=x24=-x2×-x2
 2x2-x+18=02x2-x2-x2+18=02xx-14-12x-14=0x-142x-12=0
x-14=0 or 2x-12=0x=14 or x=14
Hence, 14 is the repeated root of the given equation.

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Question 37:

10x-13=3

Answer:

Given:10x  1x = 3 10x2  1 = 3x    [Multiplying both sides by x] 10x2  3x  1 = 0 10x2  (5x  2x)  1 = 0 10x2  5x + 2x  1 = 0 5x(2x  1) + 1(2x  1) = 0 (2x  1)(5x + 1) = 0 2x  1 = 0 or 5x + 1 = 0 x = 12 or x = 15Hence, the roots of the equation are 12 and 15.

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Question 38:

2x2-5x+2=0

Answer:

Given:2x2  5x + 2 = 0 2  5x + 2x2 = 0    [Multiplying both side by x2] 2x2  5x + 2 = 0 2x2  (4x + x) + 2 = 0 2x2  4x  x + 2 = 0 2x(x  2)  1(x  2) = 0 (2x  1)(x  2) = 0 2x  1 = 0 or x  2 = 0 x = 12 or x = 2Hence, the roots of the equation are 12 and 2.

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Question 39:

Solve each of the following quadratic equations:

2x2+ax-a2=0             [CBSE 2015]

Answer:

We write, ax=2ax-ax as 2x2×-a2=-2a2x2=2ax×-ax
 2x2+ax-a2=02x2+2ax-ax-a2=02xx+a-ax+a=0x+a2x-a=0
x+a=0 or 2x-a=0x=-a or x=a2
Hence, -a and a2 are the roots of the given equation.

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Question 40:

Solve each of the following quadratic equations:

4x2+4bx-a2-b2=0                [CBSE 2015]

Answer:

We write, 4bx=2a+bx-2a-bx as 4x2×-a2-b2=-4a2-b2x2=2a+bx×-2a-bx
 4x2+4bx-a2-b2=04x2+2a+bx-2a-bx-a-ba+b=02x2x+a+b-a-b2x+a+b=02x+a+b2x-a-b=0
2x+a+b=0 or 2x-a-b=0x=-a+b2 or x=a-b2
Hence, -a+b2 and a-b2 are the roots of the given equation.

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Question 41:

Solve each of the following quadratic equations:

4x2-4a2x+a4-b4=0             [CBSE 2015]

Answer:

We write, -4a2x=-2a2+b2x-2a2-b2x as 4x2×a4-b4=4a4-b4x2=-2a2+b2x×-2a2-b2x
 4x2-4a2x+a4-b4=04x2-2a2+b2x-2a2-b2x+a2-b2a2+b2=02x2x-a2+b2-a2-b22x-a2+b2=02x-a2+b22x-a2-b2=0
2x-a2+b2=0 or 2x-a2-b2=0x=a2+b22 or x=a2-b22
Hence, a2+b22 and a2-b22 are the roots of the given equation.

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Question 42:

Solve each of the following quadratic equations:

x2+5x-a2+a-6=0                           [CBSE 2015]

Answer:

We write, 5x=a+3x-a-2x as x2×-a2+a-6=-a2+a-6x2=a+3x×-a-2x
 x2+5x-a2+a-6=0x2+a+3x-a-2x-a+3a-2=0xx+a+3-a-2x+a+3=0x+a+3x-a-2=0
x+a+3=0 or x-a-2=0x=-a+3 or x=a-2
Hence, -a+3 and a-2 are the roots of the given equation.

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Question 43:

Solve each of the following quadratic equations:

x2-2ax-4b2-a2=0                         [CBSE 2015]

Answer:

We write, -2ax=2b-ax-2b+ax as x2×-4b2-a2=-4b2-a2x2=2b-ax×-2b+ax
 x2-2ax-4b2-a2=0x2+2b-ax-2b+ax-2b-a2b+a=0xx+2b-a-2b+ax+2b-a=0x+2b-ax-2b+a=0
x+2b-a=0 or x-2b+a=0x=-2b-a or x=2b+ax=a-2b or x=a+2b
Hence, a-2b and a+2b are the roots of the given equation.

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Question 44:

Solve each of the following quadratic equations:

x2-2b-1x+b2-b-20=0                     [CBSE 2015]

Answer:

We write, -2b-1x=-b-5x-b+4x as x2×b2-b-20=b2-b-20x2=-b-5x×-b+4x
 x2-2b-1x+b2-b-20=0x2-b-5x-b+4x+b-5b+4=0xx-b-5-b+4x-b-5=0x-b-5x-b+4=0
x-b-5=0 or x-b+4=0x=b-5 or x=b+4
Hence, b-5 and b+4 are the roots of the given equation.

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Question 45:

Solve each of the following quadratic equations:

x2+6x-a2+2a-8=0                          [CBSE 2015]

Answer:

We write, 6x=a+4x-a-2x as x2×-a2+2a-8=-a2+2a-8x2=a+4x×-a-2x
 x2+6x-a2+2a-8=0x2+a+4x-a-2x-a+4a-2=0xx+a+4-a-2x+a+4=0x+a+4x-a-2=0
x+a+4=0 or x-a-2=0x=-a+4 or x=a-2
Hence, -a+4 and a-2 are the roots of the given equation.

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Question 46:

abx2+(b2-ac)x-bc=0

Answer:

Given:abx2 + (b2  ac)x  bc = 0 abx2 + b2x  acx  bc = 0 bx(ax + b)  c(ax + b) = 0 (bx  c)(ax + b) = 0 bx  c = 0 or ax + b = 0 x = cb or x = baHence, the roots of the equation are cb and ba.

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Question 47:

Solve each of the following quadratic equations:

x2-4ax-b2+4a2=0                           [CBSE 2012]

Answer:

We write, -4ax=-b+2ax+b-2ax as x2×-b2+4a2=-b2+4a2x2=-b+2ax×b-2ax
 x2-4ax-b2+4a2=0x2-b+2ax+b-2ax-b-2ab+2a=0xx-b+2a+b-2ax-b+2a=0x-b+2ax+b-2a=0
x-b+2a=0 or x+b-2a=0x=2a+b or x=-b-2ax=2a+b or x=2a-b
Hence, 2a+b and 2a-b are the roots of the given equation.

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Question 48:

4x2-2(a2+b2)x+a2b2=0

Answer:

Given:4x2  2(a2 + b2)x + a2b2 = 0 4x2  2a2x  2b2x + a2b2 = 0 2x(2x  a2)  b2(2x  a2) = 0 (2x  b2)(2x  a2) = 0 2x  b2 = 0 or 2x  a2 = 0 x = b22 or x = a22Hence, the roots of the equation are b22 and a22.

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Question 49:

12abx2-(9a2-8b2)x-6ab=0

Answer:

Given:12abx2  (9a2  8b2)x  6ab = 0 12abx2  9a2x + 8b2x  6ab = 0 3ax(4bx  3a) + 2b(4bx  3a) = 0 (3ax + 2b)(4bx  3a) = 0 3ax + 2b = 0 or 4bx  3a = 0 x = 2b3a or x = 3a4bHence, the roots of the equation are 2b3a and 3a4b.

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Question 50:

a2b2x2+b2x-a2x-1=0

Answer:

Given:a2b2x2 + b2x  a2x  1 = 0 b2x(a2x + 1)  1(a2x + 1) = 0 (b2x  1)(a2x + 1) = 0 (b2x  1) = 0 or  (a2x + 1) = 0 x = 1b2 or x = 1a2Hence, 1b2 and 1a2 are the roots of the given equation.

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Question 51:

Solve each of the following quadratic equations:

9x2-9a+bx+2a2+5ab+2b2=0                 [CBSE 2009]

Answer:

We write, -9a+bx=-32a+bx-3a+2bx as 9x2×2a2+5ab+2b2=92a2+5ab+2b2x2=-32a+bx×-3a+2bx
9x2-9a+bx+2a2+5ab+2b2=09x2-32a+bx-3a+2bx+2a+ba+2b=03x3x-2a+b-a+2b3x-2a+b=03x-2a+b3x-a+2b=0
3x-2a+b=0 or 3x-a+2b=0x=2a+b3 or x=a+2b3
Hence, 2a+b3 and a+2b3 are the roots of the given equation.

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Question 52:

Solve each of the following quadratic equations:

16x-1=15x+1, x0, -1        [CBSE 2014]

Answer:

16x-1=15x+1, x0, -116x-15x+1=116x+16-15xxx+1=1x+16x2+x=1
x2+x=x+16                     Cross multiplicationx2-16=0x+4x-4=0x+4=0 or x-4=0
x=-4 or x=4
Hence, −4 and 4 are the roots of the given equation.

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Question 53:

Solve each of the following quadratic equations:

4x-3=52x+3,   x0, -32                        [CBSE 2014]

Answer:

4x-3=52x+3,   x0, -324x-52x+3=38x+12-5xx2x+3=33x+122x2+3x=3
x+42x2+3x=12x2+3x=x+4                 Cross multiplication2x2+2x-4=0x2+x-2=0
x2+2x-x-2=0xx+2-1x+2=0x+2x-1=0x+2=0 or x-1=0
x=-2 or x=1
Hence, −2 and 1 are the roots of the given equation.

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Question 54:

Solve each of the following quadratic equations:

3x+1-12=23x-1,    x-1, 13                 [CBSE 2014]

Answer:

3x+1-12=23x-1,    x-1, 133x+1-23x-1=129x-3-2x-2x+13x-1=127x-53x2+2x-1=12
3x2+2x-1=14x-10                    Cross multiplication3x2-12x+9=0x2-4x+3=0x2-3x-x+3=0
xx-3-1x-3=0x-3x-1=0x-3=0 or x-1=0x=3 or x=1
Hence, 1 and 3 are the roots of the given equation.

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Question 55:

Solve each of the following quadratic equations:

1x-1-1x+5=67,   x1, -5              [CBSE 2010]

Answer:

1x-1-1x+5=67,   x1, -5x+5-x+1x-1x+5=676x2+4x-5=67x2+4x-5=7
x2+4x-12=0x2+6x-2x-12=0xx+6-2x+6=0x+6x-2=0
x+6=0 or x-2=0x=-6 or x=2
Hence, −6 and 2 are the roots of the given equation.

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Question 56:

Solve each of the following quadratic equations:

12a+b+2x=12a+1b+12x               [CBSE 2013]

Answer:

12a+b+2x=12a+1b+12x12a+b+2x-12x=12a+1b2x-2a-b-2x2x2a+b+2x=2a+b2ab-2a+b4x2+4ax+2bx=2a+b2ab
4x2+4ax+2bx=-2ab4x2+4ax+2bx+2ab=04xx+a+2bx+a=0x+a4x+2b=0
x+a=0 or 4x+2b=0x=-a or x=-b2
Hence, -a and -b2 are the roots of the given equation.

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Question 57:

(x+3)(x-2)-(1-x)x=174, (x=0, 2)

Answer:

Given:(x + 3)(x  2)  (1  x)x = 174 x(x + 3)  (1  x)(x  2)(x  2)x = 174 x2 + 3x  (x  2  x2 + 2x)x2  2x = 174 x2 + 3x + x2  3x + 2x2  2x = 174 2x2 + 2x2  2x = 174 8x2 + 8 = 17x2  34x        [On cross multiplying] 9x2 + 34x + 8 = 0 9x2  34x  8 = 0 9x2  36x + 2x  8 = 0 9x(x  4) + 2(x  4) = 0 (x  4)(9x + 2) = 0 x  4 = 0 or 9x + 2 = 0 x = 4 or x = 29Hence, the roots of the equation are 4 and 29.

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Question 58:

Solve each of the following quadratic equations:

3x-47+73x-4=52,   x43                   [CBSE 2010]

Answer:

3x-47+73x-4=52,   x433x-42+4973x-4=529x2-24x+16+4921x-28=529x2-24x+6521x-28=52
18x2-48x+130=105x-14018x2-153x+270=02x2-17x+30=02x2-12x-5x+30=0
2xx-6-5x-6=0x-62x-5=0x-6=0 or 2x-5=0x=6 or x=52
Hence, 6 and 52 are the roots of the given equation.



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Question 59:

Solve each of the following quadratic equations:

xx-1+x-1x=414,   x0, 1                [CBSE 2011]

Answer:

xx-1+x-1x=414,   x0, 1x2+x-12xx-1=174x2+x2-2x+1x2-x=1742x2-2x+1x2-x=174
8x2-8x+4=17x2-17x9x2-9x-4=09x2-12x+3x-4=03x3x-4+13x-4=0
3x-43x+1=03x-4=0 or 3x+1=0x=43 or x=-13
Hence, 43 and -13 are the roots of the given equation.

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Question 60:

Solve each of the following quadratic equations:

xx+1+x+1x=2415,  x0, -1             [CBSE 2014]

Answer:

xx+1+x+1x=2415,  x0, -1x2+x+12xx+1=3415x2+x2+2x+1x2+x=34152x2+2x+1x2+x=3415
30x2+30x+15=34x2+34x4x2+4x-15=04x2+10x-6x-15=02x2x+5-32x+5=0
2x+52x-3=02x+5=0 or 2x-3=0x=-52 or x=32
Hence, -52 and 32 are the roots of the given equation.

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Question 61:

Solve each of the following quadratic equations:

x-4x-5+x-6x-7=313,  x5, 7             [CBSE 2014]

Answer:

x-4x-5+x-6x-7=313,  x5, 7x-4x-7+x-5x-6x-5x-7=103x2-11x+28+x2-11x+30x2-12x+35=1032x2-22x+58x2-12x+35=103
x2-11x+29x2-12x+35=533x2-33x+87=5x2-60x+1752x2-27x+88=0
2x2-16x-11x+88=02xx-8-11x-8=0x-82x-11=0
x-8=0 or 2x-11=0x=8 or x=112
Hence, 8 and 112 are the roots of the given equation.

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Question 62:

Solve each of the following quadratic equations:

x-1x-2+x-3x-4=313,   x2, 4              [CBSE 2010]

Answer:

x-1x-2+x-3x-4=313,   x2, 4x-1x-4+x-2x-3x-2x-4=103x2-5x+4+x2-5x+6x2-6x+8=1032x2-10x+10x2-6x+8=103
x2-5x+5x2-6x+8=533x2-15x+15=5x2-30x+402x2-15x+25=02x2-10x-5x+25=0
2xx-5-5x-5=0x-52x-5=0x-5=0 or 2x-5=0x=5 or x=52
Hence, 5 and 52 are the roots of the given equation.

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Question 63:

1(x-2)+2(x-1)=6x, (x2, 1)

Answer:

Given:1(x  2) + 2(x  1) = 6x (x  1) + 2(x  2)(x  1)(x  2) = 6x 3x  5x2  3x + 2 = 6x 3x2  5x = 6x2  18x + 12        [On cross multiplying] 3x2  13x + 12 = 0 3x2  (9 + 4)x + 12 = 0 3x2  9x  4x + 12 = 0 3x(x  3)  4(x  3) = 0 (3x  4)(x  3) = 0  3x  4 = 0 or x  3 = 0 x = 43 or x = 3Hence, the roots of the equation are 43 and 3.

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Question 64:

Solve each of the following quadratic equations:

1x+1+2x+2=5x+4,   x-1, -2, -4          [CBSE 2013C]

Answer:

1x+1+2x+2=5x+4,   x-1, -2, -4x+2+2x+2x+1x+2=5x+43x+4x2+3x+2=5x+43x+4x+4=5x2+3x+2
3x2+16x+16=5x2+15x+102x2-x-6=02x2-4x+3x-6=02xx-2+3x-2=0
x-22x+3=0x-2=0 or 2x+3=0x=2 or x=-32
Hence, 2 and -32 are the roots of the given equation.

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Question 65:

Solve each of the following quadratic equations:

33x-12x+3-22x+33x-1=5,   x13, -32       [CBSE 2014]

Answer:

33x-12x+3-22x+33x-1=5,   x13, -3233x-12-22x+322x+33x-1=539x2-6x+1-24x2+12x+96x2+7x-3=527x2-18x+3-8x2-24x-186x2+7x-3=5
19x2-42x-156x2+7x-3=519x2-42x-15=30x2+35x-1511x2+77x=011xx+7=0
x=0 or x+7=0x=0 or x=-7
Hence, 0 and −7 are the roots of the given equation.

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Question 66:

Solve each of the following quadratic equations:

37x+15x-3-45x-37x+1=11,   x35, -17     [CBSE 2014]

Answer:

37x+15x-3-45x-37x+1=11,   x35, -1737x+12-45x-325x-37x+1=11349x2+14x+1-425x2-30x+935x2-16x-3=11147x2+42x+3-100x2+120x-3635x2-16x-3=11
47x2+162x-3335x2-16x-3=1147x2+162x-33=385x2-176x-33338x2-338x=0338xx-1=0
x=0 or x-1=0x=0 or x=1
Hence, 0 and 1 are the roots of the given equation.

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Question 67:

4x-32x+1-102x+14x-3=3, x-12, 34

Answer:

Given:4x  32x + 1  102x + 14x  3 = 3Putting 4x  32x + 1 = y, we get:y  10y = 3 y2  10y = 3 y2  10 = 3y [On cross multiplying] y2  3y  10 = 0 y2  (5  2)y  10 = 0 y2  5y + 2y  10 = 0 y(y  5) + 2(y  5) = 0 (y  5)(y + 2) = 0 y  5 = 0 or y + 2 = 0 y = 5 or y = 2Case IIf y = 5, we get:4x  32x + 1 = 5 4x  3 = 5(2x + 1) [On cross multiplying] 4x  3 = 10x + 5 6x = 8 6x = 8 x = 8463 x = 43Case IIIf y = 2, we get:4x  32x + 1 2 4x  3 = 2(2x + 1) 4x  3 = 4x  2 8x = 1 x = 18Hence, the roots of the equation are 43 and 18.

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Question 68:

xx+12-5xx+1+6=0, (x-1)

Answer:

Given:xx + 12  5xx + 1 + 6 = 0Putting xx + 1= y, we get:y2  5y + 6 = 0 y2  5y + 6 = 0 y2  (3 + 2)y + 6 = 0 y2  3y  2y + 6 = 0 y(y  3)  2(y  3) = 0 (y  3)(y  2) = 0 y  3 = 0 or y - 2 = 0 y = 3 or y = 2Case IIf y=3, we get:xx + 1 = 3 x = 3(x + 1) [On cross multiplying] x = 3x + 3 x = 32Case IIIf y = 2, we get:xx + 1 = 2 x = 2(x + 1) x = 2x + 2 x = 2 x = 2 Hence, the roots of the equation are 32 and 2.

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Question 69:

a(x-b)+b(x-a)=2, (xb, a)

Answer:

a(x  b) + b(x  a) = 2 [a(x  b)  1] + [b(x  a)  1] = 0 a  (x  b)x  b + b  (x  a)x  a = 0 a  x + bx  b + a  x + bx  a = 0 (a  x + b)[1(x  b) + 1(x  a)] = 0 (a  x + b)[(x  a) + (x  b)(x  b)(x  a)] = 0 (a  x + b)[2x  (a + b)(x  b)(x  a)] = 0 (a  x + b)[2x  (a + b)] = 0 a  x + b = 0 or 2x  (a + b) = 0 x = a + b or x = a + b2Hence, the roots of the equation are (a + b) and (a + b2).  

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Question 70:

a(ax-1)+b(bx-1)=(a+b), x1a,1b

Answer:

a(ax  1) + b(bx  1) = (a + b) [a(ax  1)  b] + [b(bx  1)  a] = 0 a  b(ax  1)ax  1 + b  a(bx  1)bx  1 = 0 a  abx + bax  1 + a  abx + bbx  1 = 0 (a  abx + b)[1(ax  1) + 1(bx  1)] = 0 (a  abx + b)[(bx  1) + (ax  1)(ax  1)(bx  1)] = 0 (a  abx + b)[(a + b)x  2(ax  1)(bx  1)] = 0 (a  abx + b)[(a + b)x  2] = 0 a  abx + b = 0 or (a + b)x  2 = 0 x = (a + b)ab or x = 2(a + b) Hence, the roots of the equation are (a + b)ab and 2(a + b).

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Question 71:

3(x+2)+3-x=10

Answer:

 3(x+2) + 3x = 103x.9  + 13x = 10Let 3x be equal to y. 9y + 1y = 10 9y2 + 1 = 10y 9y2 - 10y + 1 = 0 (y  1)(9y  1) = 0 y  1 = 0 or 9y  1 = 0 y = 1 or y = 19 3x = 1 or 3x = 19 3x = 30  or 3x = 32 x = 0 or x = 2Hence, 0 and 2 are the roots of the given equation.

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Question 72:

4(x+1)+4(1-x)=10

Answer:

Given:4(x+1) + 4(1x) = 10 4x.4 + 41.14x = 10Let 4xbe y. 4y + 4y = 10 4y2  10y + 4 = 0 4y2  8y  2y + 4 = 0 4y(y  2)  2(y  2) = 0 y = 2 or y = 24 = 12 4x = 2 or 12 4x = 22x = 21  or  22x = 2-1 x = 12 or  x = -12Hence, 12 and -12 are roots of the given equation.

Page No 447:

Question 73:

22x-3.2(x+2)+32=0

Answer:

Given:22x  3.2(x+2) + 32 = 0 (2x)2  3.2x.22 + 32 = 0Let 2x be y. y2  12y + 32 = 0 y2  8y  4y + 32 = 0 y(y  8)  4(y  8) = 0 (y  8) = 0 or (y  4) = 0 y = 8 or y = 4 2x = 8 or 2x = 4 2x = 23 or 2x = 22 x= 2 or 3 Hence, 2 and 3 are the roots of the given equation.



Page No 455:

Question 1:

Solve each of the following equations by using the method of completing the square:

x2-6x+3=0

Answer:

x2-6x+3=0x2-6x=-3x2-2×x×3+32=-3+32         Adding 32 on both sidesx-32=-3+9=6
x-3=±6               Taking square root on both sidesx-3=6 or x-3=-6x=3+6 or x=3-6
Hence, 3+6 and 3-6 are the roots of the given equation.

Page No 455:

Question 2:

Solve each of the following equations by using the method of completing the square:

x2-4x+1=0

Answer:

x2-4x+1=0x2-4x=-1x2-2×x×2+22=-1+22            Adding 22 on both sidesx-22=-1+4=3
x-2=±3                Taking square root on both sidesx-2=3 or x-2=-3x=2+3 or x=2-3
Hence, 2+3 and 2-3 are the roots of the given equation.

Page No 455:

Question 3:

Solve each of the following equations by using the method of completing the square:

x2+8x-2=0

Answer:

x2+8x-2=0x2+8x=2x2+2×x×4+42=2+42           Adding 42 on both sidesx+42=2+16=18
x+4=±18=±32                 Taking square root on both sidesx+4=32 or x+4=-32x=-4+32 or x=-4-32
Hence, -4+32 and -4-32 are the roots of the given equation.

Page No 455:

Question 4:

Solve each of the following equations by using the method of completing the square:

4x2+43x+3=0

Answer:

4x2+43x+3=04x2+43x=-32x2+2×2x×3+32=-3+32          Adding 32 on both sides 2x+32=-3+3=0
2x+3=0x=-32
Hence, -32 is the repeated root of the given equation.

Page No 455:

Question 5:

Solve each of the following equations by using the method of completing the square:

2x2+5x-3=0

Answer:

2x2+5x-3=04x2+10x-6=0         Multiplying both sides by 24x2+10x=62x2+2×2x×52+522=6+522            Adding 522 on both sides
2x+522=6+254=24+254=494=7222x+52=±72            Taking square root on both sides2x+52=72 or 2x+52=-722x=72-52=22=1 or 2x=-72-52=-122=-6
x=12 or x=-3
Hence, 12 and -3 are the roots of the given equation.

Page No 455:

Question 6:

Solve each of the following equations by using the method of completing the square:

3x2-x-2=0

Answer:

3x2-x-2=09x2-3x-6=0                Multiplying both sides by 39x2-3x=63x2-2×3x×12+122=6+122          Adding 122 on both sides
3x-122=6+14=254=5223x-12=±52                       Taking square root on both sides3x-12=52 or 3x-12=-523x=52+12=62=3 or 3x=-52+12=-42=-2
x=1 or x=-23
Hence, 1 and -23 are the roots of the given equation.

Page No 455:

Question 7:

Solve each of the following equations by using the method of completing the square:

8x2-14x-15=0

Answer:

8x2-14x-15=016x2-28x-30=0                Multiplying both sides by 216x2-28x=304x2-2×4x×72+722=30+722        Adding 722 on both sides
4x-722=30+494=1694=13224x-72=±132                  Taking square root on both sides4x-72=132 or 4x-72=-1324x=132+72=202=10 or 4x=-132+72=-62=-3
x=52 or x=-34
Hence, 52 and -34 are the roots of the given equation.

Page No 455:

Question 8:

Solve each of the following equations by using the method of completing the square:

7x2+3x-4=0

Answer:

7x2+3x-4=049x2+21x-28=0                 Multiplying both sides by 749x2+21x=287x2+2×7x×32+322=28+322          Adding 322 on both sides 
7x+322=28+94=1214=11227x+32=±112               Taking square root on both sides7x+32=112 or 7x+32=-1127x=112-32=82=4 or 7x=-112-32=-142=-7
x=47 or x=-1
Hence, 47 and −1 are the roots of the given equation.

Page No 455:

Question 9:

Solve each of the following equations by using the method of completing the square:

3x2-2x-1=0

Answer:

3x2-2x-1=09x2-6x-3=0                 Multiplying both sides by 39x2-6x=33x2-2×3x×1+12=3+12           Adding 12 on both sides
3x-12=3+1=4=223x-1=±2                 Taking square root on both sides3x-1=2 or 3x-1=-23x=3 or 3x=-1
x=1 or x=-13
Hence, 1 and -13 are the roots of the given equation.

Page No 455:

Question 10:

Solve each of the following equations by using the method of completing the square:

5x2-6x-2=0

Answer:

5x2-6x-2=025x2-30x-10=0                  Multiplying both sides by 525x2-30x=105x2-2×5x×3+32=10+32           Adding 32 on both sides
5x-32=10+9=195x-3=±19              Taking square root on both sides5x-3=19 or 5x-3=-195x=3+19 or 5x=3-19
x=3+195 or x=3-195
Hence, 3+195 and 3-195 are the roots of the given equation.

Page No 455:

Question 11:

Solve each of the following equations by using the method of completing the square:

2x2-5x+2=0

Answer:

2x2-5x+2=02-5x+2x2x2=02x2-5x+2=04x2-10x+4=0             Multiplying both sides by 2
4x2-10x=-42x2-2×2x×52+522=-4+522         Adding 522 on both sides2x-522=-4+254=94=3222x-52=±32                  Taking square root on both sides
2x-52=32 or 2x-52=-322x=32+52=82=4 or 2x=-32+52=22=1x=2 or x=12
Hence, 2 and 12 are the roots of the given equation.

Page No 455:

Question 12:

Solve each of the following equations by using the method of completing the square:

4x2+4bx-a2-b2=0

Answer:

4x2+4bx-a2-b2=04x2+4bx=a2-b22x2+2×2x×b+b2=a2-b2+b2                Adding b2 on both sides2x+b2=a2
2x+b=±a                Taking square root on both sides2x+b=a or 2x+b=-a2x=a-b or 2x=-a-bx=a-b2 or x=-a+b2
Hence, a-b2 and -a+b2 are the roots of the given equation.

Page No 455:

Question 13:

Solve each of the following equations by using the method of completing the square:

x2-2+1x+2=0

Answer:

x2-2+1x+2=0x2-2+1x=-2x2-2×x×2+12+2+122=-2+2+122               Adding 2+122 on both sides
x-2+122=-42+2+1+224=2-22+14=2-122x-2+12=±2-12                 Taking square root on both sidesx-2+12=2-12 or x-2+12=-2-12x=2+12+2-12 or x=2+12-2-12
x=222=2 or x=22=1
Hence, 2 and 1 are the roots of the given equation.

Page No 455:

Question 14:

Solve each of the following equations by using the method of completing the square:

2x2-3x-22=0

Answer:

2x2-3x-22=02x2-32x-4=0              Multiplying both sides by 22x2-32x=42x2-2×2x×32+322=4+322            Adding 322 on both sides
2x-322=4+94=254=5222x-32=±52                     Taking square root on both sides2x-32=52 or 2x-32=-522x=52+32=82=4 or 2x=-52+32=-22=-1
x=42=22 or x=-12=-22
Hence, 22 and -22 are the roots of the given equation.

Page No 455:

Question 15:

Solve each of the following equations by using the method of completing the square:

3x2+10x+73=0

Answer:

3x2+10x+73=03x2+103x+21=0                 Multiplying both sides by 33x2+103x=-213x2+2×3x×5+52=-21+52         Adding 52 on both sides
3x+52=-21+25=4=223x+5=±2                 Taking square root on both sides3x+5=2 or 3x+5=-23x=-3 or 3x=-7
x=-33=-3 or x=-73=-733
Hence, -3 and -733 are the roots of the given equation.

Page No 455:

Question 16:

By using the method of completing the square, show that the equation 2x2+x+4=0 has no real roots.

Answer:

2x2+x+4=04x2+2x+8=0                 Multiplying both sides by 24x2+2x=-82x2+2×2x×12+122=-8+122         Adding 122 on both sides
2x+122=-8+14=-314<0
But, 2x+122 cannot be negative for any real value of x.
So, there is no real value of x satisfying the given equation.

Hence, the given equation has no real roots.



Page No 460:

Question 1:

Find the discriminant of each of the following equations:

(i) 2x2-7x+6=0
(ii) 3x2-2x+8=0 
(iii) 2x2-52x+4=0
(iv) 3x2+22x-23=0
(v) x-12x-1=0
(vi) 1-x=2x22x27x+6=02x27x+6=0

Answer:

(i) 2x2  7x + 6 = 0Here,a = 2,b = 7,c = 6Discriminant D is diven by:D = b2  4ac= (7)2  4 × 2 × 6= 49  48= 1

(ii) 3x2  2x + 8 = 0Here,a = 3,b = 2,c = 8Discriminant D is given by:D = b2  4ac(2)2  4 × 3 × 84  96= 92

(iii) 2x2  52x + 4 = 0Here,a = 2,b = 52,c = 4Discriminant D is given by:D = b2  4ac= (52)2  4 × 2 × 4= (25 × 2)  32= 50  32= 18

(iv)3x2 + 22x  23 = 0Here,a = 3,b = 22,c = 23Discriminant D is given by:D = b2  4ac= (22)2  4 × 3 × (23)= (4 × 2) + (8 × 3)= 8 + 24= 32

(v) x-12x-1=0

2x2-3x+1=0

Comparing it with ax2+bx+c=0, we get

a = 2, b = −3 and c = 1

∴ Discriminant, D = b2-4ac=-32-4×2×1=9-8=1

(vi) 1  x = 2x2 2x2 + x  1 = 0Here,a = 2,b = 1,c = 1Discriminant D is given by:D = b2  4ac= 12  4 × 2(1)= 1 + 8= 9

Page No 460:

Question 2:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

x2-4x-1=0

Answer:

Given:x2  4x  1 = 0On comparing it with ax2 + bx + c = 0, we get:a = 1, b = 4 and c = 1Discriminant D is given by:D = (b2  4ac)(4)2  4 × 1 × (1)= 16 + 420= 20 > 0Hence, the roots of the equation are real.Roots α and β are given by:α = b + D2a = (4) + 202 × 1 = 4 + 252 = 2(2 + 5)2 = (2 + 5)β = b  D2a = (4)  202 = 4  252 = 2(2  5)2 = (2  5)Thus, the roots of the equation are (2 + 5)  and (2  5).

Page No 460:

Question 3:

x2-6x+4=0

Answer:

Given:x2  6x + 4 = 0On comparing it with ax2 + bx + c = 0, we get:a = 1, b = 6 and c = 4Discriminant D is given by:D = (b2  4ac)(6)2  4 × 1 × 4= 36  16= 20 > 0Hence, the roots of the equation are real.Roots α and β are given by:α = b + D2a = (6) + 202 × 1 = 6 + 252 = 2(3 + 5)2 (3 + 5)β = b  D2a =(6)  202 × 1 = 6  252 = 2(3  5)2 = (3  5)Thus, the roots of the equation are (3 + 25) and (3  25).

Page No 460:

Question 4:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

2x2+x-4=0

Answer:

The given equation is 2x2+x-4=0.

Comparing it with ax2+bx+c=0, we get

a = 2, b = 1 and c = −4

∴ Discriminant, D = b2-4ac=12-4×2×-4=1+32=33>0

So, the given equation has real roots.

Now, D=33

 α=-b+D2a=-1+332×2=-1+334β=-b-D2a=-1-332×2=-1-334

Hence, -1+334 and -1-334 are the roots of the given equation.

Page No 460:

Question 5:

25x2+30x+7=0

Answer:

Given:25x2 + 30x + 7 = 0On comparing it with ax2 + bx + c = 0, we get:a = 25, b = 30 and c = 7Discriminant D is given by:D = (b2  4ac)= 302  4 × 25 × 7= 900  700= 200= 200 > 0Hence, the roots of the equation are real.Roots α and β are given by:α = b + D2a = 30 + 2002 × 25 = 30 + 10250 = 10(3 + 2)50 = (3 + 2)5β = b  D2a = 30  2002 × 25 = 30  10250 = 10(3  2)50 = (3  2)5Thus, the roots of the equation are (3 + 2)5 and (3  2)5.

Page No 460:

Question 6:

16x2=24x+1

Answer:

Given:16x2 = 24x + 1⇒ 16x2  24x  1 = 0On comparing it with ax2 + bx + c = 0a = 16, b = 24 and c = 1Discriminant D is given by:D = (b2  4ac)(24)2  4 × 16 × (1)= 576 + (64)= 640 > 0Hence, the roots of the equation are real,Roots α and β are given by:α = b + D2a (24) + 6402 × 16 = 24 + 81032 = 8(3 + 10)32 = (3 + 10)4β = b  D2a = (24)  6402 × 16 = 24  81032 = 8(3  10)32 = (3  10)4Thus, the roots of the equation are (3 + 10)4 and  (3  10)4.

Page No 460:

Question 7:

15x2-28=x

Answer:

Given:15x2  28 = x⇒ 15x2  x  28 = 0On comparing it with ax2 + bx + c = 0, we get:a = 15, b = 1 and c = 28Discriminant D is given by:D = (b2  4ac)(1)2  4 × 15 × (28)= 1  (1680)= 1 + 1680= 1681= 1681 > 0Hence, the roots of the equation are real.Roots α and β are given by:α = b + D2a = (1) + 16812 × 15 = 1 + 4130 = 4230 = 75β = b  D2a = (1)  16812 × 15 = 1  4130 = 4030 = 43Thus, the roots of the equation are 75 and 43.

Page No 460:

Question 8:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

2x2-22x+1=0

Answer:

The given equation is 2x2-22x+1=0.

Comparing it with ax2+bx+c=0, we get

a = 2, b-22 and c = 1

∴ Discriminant, D = b2-4ac=-222-4×2×1=8-8=0

So, the given equation has real roots.

Now, D=0

 α=-b+D2a=--22+02×2=224=22β=-b-D2a=--22-02×2=224=22

Hence, 22 is the repeated root of the given equation.

Page No 460:

Question 9:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

2x2+7x+52=0                      [CBSE 2013]

Answer:

The given equation is 2x2+7x+52=0.

Comparing it with ax2+bx+c=0, we get

a = 2, b = 7 and c = 52

∴ Discriminant, D = b2-4ac=72-4×2×52=49-40=9>0

So, the given equation has real roots.

Now, D=9=3

 α=-b+D2a=-7+32×2=-422=-2β=-b-D2a=-7-32×2=-1022=-522

Hence, -2 and -522 are the roots of the given equation.

Page No 460:

Question 10:

3x2+10x-83=0

Answer:

Given:3x2 + 10x  83 = 0On comparing it with ax2 + bx + c = 0, we get:a = 3, b = 10 and c = 83Discriminant D is given by:D = (b2  4ac)= (10)2  4 × 3 × (83)= 100 + 96= 196 > 0Hence, the roots of the equation are real.Roots α and β are given by: α b + D2a = 10 + 19623 = 10 + 1423 = 423 = 23 = 23 × 33 = 233β = b  D2a = 10  19623 = 10  1423 = 2423 = 123 = 123 × 33 = 1233 = 43Thus, the roots of the equation are 233 and 43.

Page No 460:

Question 11:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

3x2-22x-23=0                        [CBSE 2015]

Answer:

The given equation is 3x2-22x-23=0.

Comparing it with ax2+bx+c=0, we get

a = 3, b-22 and c = -23

∴ Discriminant, D = b2-4ac=-222-4×3×-23=8+24=32>0

So, the given equation has real roots.

Now, D=32=42

 α=-b+D2a=--22+422×3=6223=6β=-b-D2a=--22-422×3=-2223=-63

Hence, 6 and -63 are the roots of the given equation.

Page No 460:

Question 12:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

2x2+63x-60=0                                           [CBSE 2011, 15]

Answer:

The given equation is 2x2+63x-60=0.

Comparing it with ax2+bx+c=0, we get

a = 2, b63 and c = -60

∴ Discriminant, D = b2-4ac=632-4×2×-60=108+480=588>0

So, the given equation has real roots.

Now, D=588=143

 α=-b+D2a=-63+1432×2=834=23β=-b-D2a=-63-1432×2=-2034=-53

Hence, 23 and -53 are the roots of the given equation.

Page No 460:

Question 13:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

43x2+5x-23=0                                           [CBSE 2013]

Answer:

The given equation is 43x2+5x-23=0.

Comparing it with ax2+bx+c=0, we get

a = 43, b = 5 and c = -23

∴ Discriminant, D = b2-4ac=52-4×43×-23=25+96=121>0

So, the given equation has real roots.

Now, D=121=11

 α=-b+D2a=-5+112×43=683=34β=-b-D2a=-5-112×43=-1683=-233

Hence, 34 and -233 are the roots of the given equation.

Page No 460:

Question 14:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

3x2-26x+2=0                                           [CBSE 2012]

Answer:

The given equation is 3x2-26x+2=0.

Comparing it with ax2+bx+c=0, we get

a = 3, b-26 and c = 2

∴ Discriminant, D = b2-4ac=-262-4×3×2=24-24=0

So, the given equation has real roots.

Now, D=0

 α=-b+D2a=--26+02×3=266=63β=-b-D2a=--26-02×3=266=63

Hence, 63 is the repeated root of the given equation.



Page No 461:

Question 15:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

23x2-5x+3=0                                           [CBSE 2011]

Answer:

The given equation is 23x2-5x+3=0.

Comparing it with ax2+bx+c=0, we get

a = 23, b-5 and c = 3

∴ Discriminant, D = b2-4ac=-52-4×23×3=25-24=1>0

So, the given equation has real roots.

Now, D=1=1

 α=-b+D2a=--5+12×23=643=32β=-b-D2a=--5-12×23=443=33

Hence, 32 and 33 are the roots of the given equation.

Page No 461:

Question 16:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

x2+x+2=0

Answer:

The given equation is x2+x+2=0.

Comparing it with ax2+bx+c=0, we get

a = 1, b = 1 and c = 2

∴ Discriminant, D = b2-4ac=12-4×1×2=1-8=-7<0

Hence, the given equation has no real roots (or real roots does not exist).

Page No 461:

Question 17:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

2x2+ax-a2=0                                           [CBSE 2015]

Answer:

The given equation is 2x2+ax-a2=0.

Comparing it with Ax2+Bx+C=0, we get

A = 2, B = a and C = -a2

∴ Discriminant, D = B2-4AC=a2-4×2×-a2=a2+8a2=9a20

So, the given equation has real roots.

Now, D=9a2=3a

 α=-B+D2A=-a+3a2×2=2a4=a2β=-B-D2A=-a-3a2×2=-4a4=-a

Hence, a2 and -a are the roots of the given equation.

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Question 18:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

x2-3+1x+3=0                                           [CBSE 2015]

Answer:

The given equation is x2-3+1x+3=0.

Comparing it with ax2+bx+c=0, we get

a = 1, b-3+1 and c = 3

∴ Discriminant, D = b2-4ac=-3+12-4×1×3=3+1+23-43=3-23+1=3-12>0

So, the given equation has real roots.

Now, D=3-12=3-1

 α=-b+D2a=--3+1+3-12×1=3+1+3-12=232=3β=-b-D2a=--3+1-3-12×1=3+1-3+12=22=1

Hence, 3 and 1 are the roots of the given equation.

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Question 19:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

2x2+53x+6=0

Answer:

The given equation is 2x2+53x+6=0.

Comparing it with ax2+bx+c=0, we get

a = 2, b53 and c = 6

∴ Discriminant, D = b2-4ac=532-4×2×6=75-48=27>0

So, the given equation has real roots.

Now, D=27=33

 α=-b+D2a=-53+332×2=-234=-32β=-b-D2a=-53-332×2=-834=-23

Hence, -32 and -23 are the roots of the given equation.

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Question 20:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

3x2-2x+2=0

Answer:

The given equation is 3x2-2x+2=0.

Comparing it with ax2+bx+c=0, we get

a = 3, b = −2 and c = 2

∴ Discriminant, D = b2-4ac=-22-4×3×2=4-24=-20<0

Hence, the given equation has no real roots (or real roots does not exist).

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Question 21:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

x+1x=3,  x0

Answer:

The given equation is

x+1x=3,  x0x2+1x=3x2+1=3xx2-3x+1=0

This equation is of the form ax2+bx+c=0, where a = 1, b = −3 and c = 1.

∴ Discriminant, D = b2-4ac=-32-4×1×1=9-4=5>0

So, the given equation has real roots.

Now, D=5

 α=-b+D2a=--3+52×1=3+52β=-b-D2a=--3-52×1=3-52

Hence, 3+52 and 3-52 are the roots of the given equation.

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Question 22:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

1x-1x-2=3,  x0, 2              [CBSE 2010]

Answer:

The given equation is

1x-1x-2=3,  x0, 2x-2-xxx-2=3-2x2-2x=3-2=3x2-6x

3x2-6x+2=0

This equation is of the form ax2+bx+c=0, where a = 3, b = −6 and c = 2.

∴ Discriminant, D = b2-4ac=-62-4×3×2=36-24=12>0

So, the given equation has real roots.

Now, D=12=23

 α=-b+D2a=--6+232×3=6+236=3+33β=-b-D2a=--6-232×3=6-236=3-33

Hence, 3+33 and 3-33 are the roots of the given equation.

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Question 23:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

x-1x=3,  x0                  [CBSE 2010]

Answer:

The given equation is

x-1x=3,  x0x2-1x=3x2-1=3xx2-3x-1=0

This equation is of the form ax2+bx+c=0, where a = 1, b = −3 and c = −1.

∴ Discriminant, D = b2-4ac=-32-4×1×-1=9+4=13>0

So, the given equation has real roots.

Now, D=13

 α=-b+D2a=--3+132×1=3+132β=-b-D2a=--3-132×1=3-132

Hence, 3+132 and 3-132 are the roots of the given equation.

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Question 24:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

mnx2+nm=1-2x

Answer:

The given equation is

mnx2+nm=1-2xm2x2+n2mn=1-2xm2x2+n2=mn-2mnxm2x2+2mnx+n2-mn=0

This equation is of the form ax2+bx+c=0, where a = m2, b = 2mn and c = n2-mn.

∴ Discriminant, D = b2-4ac=2mn2-4×m2×n2-mn=4m2n2-4m2n2+4m3n=4m3n>0

So, the given equation has real roots.

Now, D=4m3n=2mmn

 α=-b+D2a=-2mn+2mmn2×m2=2m-n+mn2m2=-n+mnmβ=-b-D2a=-2mn-2mmn2×m2=-2mn+mn2m2=-n-mnm

Hence, -n+mnm and -n-mnm are the roots of the given equation.

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Question 25:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

36x2-12ax+a2-b2=0

Answer:

The given equation is 36x2-12ax+a2-b2=0.

Comparing it with Ax2+Bx+C=0, we get

A = 36, B-12a and C = a2-b2

∴ Discriminant, D = B2-4AC=-12a2-4×36×a2-b2=144a2-144a2+144b2=144b2>0

So, the given equation has real roots.

Now, D=144b2=12b

 α=-B+D2A=--12a+12b2×36=12a+b72=a+b6β=-B-D2A=--12a-12b2×36=12a-b72=a-b6

Hence, a+b6 and a-b6 are the roots of the given equation.

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Question 26:

x2-2ax+(a2-b2)=0

Answer:

Given: x2  2ax + (a2  b2) = 0On comparing it with Ax2 + Bx + C = 0, we get:A = 1, B = 2a and C = (a2  b2)Discriminant D is given by: D B2  4AC    (2a)2  4 × 1 × (a2  b2)= 4a2  4a2 + 4b2  = 4b2 > 0Hence, the roots of the equation are real.Roots α and β are given by:α b + D2a = (2a) + 4b22 × 1 = 2a + 2b2 = 2(a + b)2 = (a + b)β b  D2a = (2a)  4b22 × 1 = 2a  2b2 = 2(a  b)2 = (a  b)Hence, the roots of the equation are (a + b) and (ab).

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Question 27:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

x2-2ax-4b2-a2=0               [CBSE 2015]

Answer:

The given equation is x2-2ax-4b2-a2=0.

Comparing it with Ax2+Bx+C=0, we get

A = 1, B-2a and C = -4b2-a2

∴ Discriminant, D = B2-4AC=-2a2-4×1×-4b2-a2=4a2+16b2-4a2=16b2>0

So, the given equation has real roots.

Now, D=16b2=4b

 α=-B+D2A=--2a+4b2×1=2a+2b2=a+2bβ=-B-D2A=--2a-4b2×1=2a-2b2=a-2b

Hence, a+2b and a-2b are the roots of the given equation.

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Question 28:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

x2+6x-a2+2a-8=0                  [CBSE 2015]

Answer:

The given equation is x2+6x-a2+2a-8=0.

Comparing it with Ax2+Bx+C=0, we get

A = 1, B = 6 and C = -a2+2a-8

∴ Discriminant, D = B2-4AC=62-4×1×-a2+2a-8=36+4a2+8a-32=4a2+8a+4=4a2+2a+1=4a+12>0

So, the given equation has real roots.

Now, D=4a+12=2a+1

 α=-B+D2A=-6+2a+12×1=2a-42=a-2β=-B-D2A=-6-2a+12×1=-2a-82=-a-4=-a+4

Hence, a-2 and -a+4 are the roots of the given equation.

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Question 29:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

x2+5x-a2+a-6=0                  [CBSE 2015]

Answer:

The given equation is x2+5x-a2+a-6=0.

Comparing it with Ax2+Bx+C=0, we get

A = 1, B = 5 and C = -a2+a-6

∴ Discriminant, D = B2-4AC=52-4×1×-a2+a-6=25+4a2+4a-24=4a2+4a+1=2a+12>0

So, the given equation has real roots.

Now, D=2a+12=2a+1

 α=-B+D2A=-5+2a+12×1=2a-42=a-2β=-B-D2A=-5-2a+12×1=-2a-62=-a-3=-a+3

Hence, a-2 and -a+3 are the roots of the given equation.

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Question 30:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

x2-4ax-b2+4a2=0                  [CBSE 2012]

Answer:

The given equation is x2-4ax-b2+4a2=0.

Comparing it with Ax2+Bx+C=0, we get

A = 1, B = −4a and C = -b2+4a2

∴ Discriminant, D = B2-4AC=-4a2-4×1×-b2+4a2=16a2+4b2-16a2=4b2>0

So, the given equation has real roots.

Now, D=4b2=2b

 α=-B+D2A=--4a+2b2×1=4a+2b2=2a+bβ=-B-D2A=--4a-2b2×1=4a-2b2=2a-b

Hence, 2a+b and 2a-b are the roots of the given equation.

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Question 31:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

4x2-4a2x+a4-b4=0           [CBSE 2015]

Answer:

The given equation is 4x2-4a2x+a4-b4=0.

Comparing it with Ax2+Bx+C=0, we get

A = 4, B = −4a2 and C = a4-b4

∴ Discriminant, D = B2-4AC=-4a22-4×4×a4-b4=16a4-16a4+16b4=16b4>0

So, the given equation has real roots.

Now, D=16b4=4b2

 α=-B+D2A=--4a2+4b22×4=4a2+b28=a2+b22β=-B-D2A=--4a2-4b22×4=4a2-b28=a2-b22

Hence, 12a2+b2 and 12a2-b2 are the roots of the given equation.

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Question 32:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

4x2+4bx-a2-b2=0              [CBSE 2015]

Answer:

The given equation is 4x2+4bx-a2-b2=0.

Comparing it with Ax2+Bx+C=0, we get

A = 4, B = 4b and C = -a2-b2

∴ Discriminant, D = B2-4AC=4b2-4×4×-a2-b2=16b2+16a2-16b2=16a2>0

So, the given equation has real roots.

Now, D=16a2=4a

 α=-B+D2A=-4b+4a2×4=4a-b8=a-b2β=-B-D2A=-4b-4a2×4=-4a+b8=-a+b2

Hence, 12a-b and -12a+b are the roots of the given equation.

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Question 33:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

x2-2b-1x+b2-b-20=0               [CBSE 2015]

Answer:

The given equation is x2-2b-1x+b2-b-20=0.

Comparing it with Ax2+Bx+C=0, we get

A = 1, B = -2b-1 and C = b2-b-20

∴ Discriminant, D = B2-4AC=-2b-12-4×1×b2-b-20=4b2-4b+1-4b2+4b+80=81>0

So, the given equation has real roots.

Now, D=81=9

 α=-B+D2A=--2b-1+92×1=2b+82=b+4β=-B-D2A=--2b-1-92×1=2b-102=b-5

Hence, b+4 and b-5 are the roots of the given equation.

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Question 34:

3a2x2+8abx+4b2=0, a0

Answer:

Given:3a2x2 + 8abx + 4b2 = 0   On comparing it with Ax2 + Bx + C = 0, we get:A = 3a2, B = 8ab and C = 4b2   Discriminant D is given by:D = (B2  4AC)             = (8ab)2  4 × 3a2 × 4b2     = 16a2b2 > 0Hence, the roots of the equation are real.Roots α and β are given by:α b + D2a = 8ab + 16a2b22 × 3a2 = 8ab + 4ab6a2 = 4ab6a2 = 2b3aβ = b  D2a = 8ab  16a2b22 × 3a2 = 8ab  4ab6a2 = 12ab6a2 = 2baThus, the roots of the equation are 2b3a and 2ba.

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Question 35:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

a2b2x2-4b4-3a4x-12a2b2=0, a0 and b0                  [CBSE 2006]

Answer:

The given equation is a2b2x2-4b4-3a4x-12a2b2=0.

Comparing it with Ax2+Bx+C=0, we get

A = a2b2, B = -4b4-3a4 and C = -12a2b2

∴ Discriminant, D = B2-4AC=-4b4-3a42-4×a2b2×-12a2b2=16b8-24a4b4+9a8+48a4b4=16b8+24a4b4+9a8=4b4+3a42>0

So, the given equation has real roots.

Now, D=4b4+3a42=4b4+3a4

 α=-B+D2A=--4b4-3a4+4b4+3a42×a2b2=8b42a2b2=4b2a2β=-B-D2A==--4b4-3a4-4b4+3a42×a2b2=-6a42a2b2=-3a2b2

Hence, 4b2a2 and -3a2b2 are the roots of the given equation.

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Question 36:

12abx2-(9a2-8b2)x-6ab=0, where a0 and b0

Answer:

Given:12abx2  (9a2  8b2)x  6ab = 0     On comparing it with Ax2 + Bx + C = 0, we get:     = 12ab, B = (9a2  8b2) and C = 6abDiscriminant D is given by: D B2  4AC  = [(9a2  8b2)]2  4 × 12ab × (6ab)     = 81a4  144a2b2 + 64b4 + 288a2b2   = 81a4 + 144a2b2 + 64b4   = (9a2 + 8b2)2 > 0Hence, the roots of the equation are equal.Roots α and β are given by:α B + D2A = [(9a2  8b2)] + (9a2 + 8b2)22 × 12ab = 9a2  8b2 + 9a2 + 8b224ab = 18a224ab = 3a4bβ B  D2A = [(9a2  8b2)]  (9a2+8b2)22 × 12ab = 9a2  8b2  9a2  8b224ab = 16b224ab = 2b3aThus, the roots of the equation are 3a4b and 2b3a.



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Question 1:

Find the nature of the roots of the following quadratic equations:
(i) 2x2-8x+5=0
(ii) 3x2-26x+2=0
(iii) 5x2-4x+1=0
(iv) 5xx-2+6=0
(v) 12x2-415x+5=0
(vi) x2-x+2=0

Answer:

(i) The given equation is 2x2-8x+5=0.

This is of the form ax2+bx+c=0, where a = 2, b = −8 and c = 5.

∴ Discriminant, D = b2-4ac=-82-4×2×5=64-40=24>0

Hence, the given equation has real and unequal roots.

(ii) The given equation is 3x2-26x+2=0.

This is of the form ax2+bx+c=0, where a = 3, b-26 and c = 2.

∴ Discriminant, D = b2-4ac=-262-4×3×2=24-24=0

Hence, the given equation has real and equal roots.

(iii) The given equation is 5x2-4x+1=0.

This is of the form ax2+bx+c=0, where a = 5, b = −4 and c = 1.

∴ Discriminant, D = b2-4ac=-42-4×5×1=16-20=-4<0

Hence, the given equation has no real roots.

(iv) The given equation is

5xx-2+6=05x2-10x+6=0

This is of the form ax2+bx+c=0, where a = 5, b = −10 and c = 6.

∴ Discriminant, D = b2-4ac=-102-4×5×6=100-120=-20<0

Hence, the given equation has no real roots.

(v) The given equation is 12x2-415x+5=0.

This is of the form ax2+bx+c=0, where a = 12, b-415 and c = 5.

∴ Discriminant, D = b2-4ac=-4152-4×12×5=240-240=0

Hence, the given equation has real and equal roots.

(vi) The given equation is x2-x+2=0.

This is of the form ax2+bx+c=0, where a = 1, b = −1 and c = 2.

∴ Discriminant, D = b2-4ac=-12-4×1×2=1-8=-7<0

Hence, the given equation has no real roots.

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Question 2:

If a and b are distinct real numbers, show that the quadratic equation 2a2+b2x2+2a+bx+1=0 has no real roots.

Answer:

The given equation is 2a2+b2x2+2a+bx+1=0.
D=2a+b2-4×2a2+b2×1       =4a2+2ab+b2-8a2+b2       =4a2+8ab+4b2-8a2-8b2       =-4a2+8ab-4b2       =-4a2-2ab+b2       =-4a-b2<0
Hence, the given equation has no real roots.

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Question 3:

Show that the roots of the equation x2+px-q2=0 are real for all real value of p and q.

Answer:

Given: x2 + px  q2 = 0Here, a = 1, b = p and c = q2Discriminant D is given by:D = (b2  4ac)= p2  4 × 1 × (q2)= (p2 + 4q2) > 0D>0 for all real values of p and q.Thus, the roots of the equation are real.

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Question 4:

For what values of k are the roots of the quadratic equation 3x2+2kx+27=0 real and equal?

Answer:

Given: 3x2 + 2kx + 27 = 0Here, a = 3, b = 2k and c = 27It is given that the roots of the equation are real and equal; therefore, we have:D = 0 (2k)2  4 × 3 × 27 = 0 4k2  324 = 0 4k2 = 324 k2 = 81 k = ±9 k = 9 or k = 9

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Question 5:

For what value of k are the roots of the quadratic equation kxx-25+10=0 real and equal?                                         [CBSE 2013]

Answer:

The given equation is

kxx-25+10=0kx2-25kx+10=0

This is of the form ax2+bx+c=0, where a = k, b = -25k and c = 10.

D=b2-4ac=-25k2-4×k×10=20k2-40k

The given equation will have real and equal roots if D = 0.

20k2-40k=020kk-2=0k=0 or k-2=0k=0 or k=2

But, for k = 0, we get 10 = 0, which is not true.

Hence, 2 is the required value of k.

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Question 6:

For what values of p are the roots of the equation 4x2+px+3=0 real and equal?                                                             [CBSE 2014]

Answer:

The given equation is 4x2+px+3=0.

This is of the form ax2+bx+c=0, where a = 4, b = p and c = 3.

D=b2-4ac=p2-4×4×3=p2-48

The given equation will have real and equal roots if D = 0.

p2-48=0p2=48p=±48=±43

Hence, 43 and -43 are the required values of p.

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Question 7:

Find the non-zero value of k for which the roots of the quadratic equation 9x2-3kx+k=0 are real and equal.                             [CBSE 2014]

Answer:

The given equation is 9x2-3kx+k=0.

This is of the form ax2+bx+c=0, where a = 9, b = −3k and c = k.

D=b2-4ac=-3k2-4×9×k=9k2-36k

The given equation will have real and equal roots if D = 0.

9k2-36k=09kk-4=0k=0 or k-4=0k=0 or k=4

But, k ≠ 0        (Given)

Hence, the required value of k is 4.

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Question 8:

Find the values of k for which the quadratic equation 3k+1x2+2k+1x+1=0 has real and equal roots.                                  [CBSE 2014]

Answer:

The given equation is 3k+1x2+2k+1x+1=0.

This is of the form ax2+bx+c=0, where a = 3k +1, b = 2(k + 1) and c = 1.

D=b2-4ac       =2k+12-4×3k+1×1       =4k2+2k+1-43k+1       =4k2+8k+4-12k-4
        =4k2-4k

The given equation will have real and equal roots if D = 0.

4k2-4k=04kk-1=0k=0 or k-1=0k=0 or k=1

Hence, 0 and 1 are the required values of k.

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Question 9:

Find the values of p for which the quadratic equation 2p+1x2-7p+2x+7p-3=0 has real and equal roots.                           [CBSE 2014]

Answer:

The given equation is 2p+1x2-7p+2x+7p-3=0.

This is of the form ax2+bx+c=0, where a = 2p +1, b = (7p + 2) and c = 7p − 3.

D=b2-4ac       =-7p+22-4×2p+1×7p-3       =49p2+28p+4-414p2+p-3       =49p2+28p+4-56p2-4p+12       
       =-7p2+24p+16

The given equation will have real and equal roots if D = 0.

-7p2+24p+16=07p2-24p-16=07p2-28p+4p-16=07pp-4+4p-4=0
p-47p+4=0p-4=0 or 7p+4=0p=4 or p=-47

Hence, 4 and -47 are the required values of p.

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Question 10:

Find the values of p for which the quadratic equation p+1x2-6p+1x+3p+9=0, p-1 has equal roots. Hence, find the roots of the equation.                                                                                                                                                                       [CBSE 2015]

Answer:

The given equation is p+1x2-6p+1x+3p+9=0.

This is of the form ax2+bx+c=0, where a = p +1, b = 6(p + 1) and c = 3(p + 9).

D=b2-4ac       =-6p+12-4×p+1×3p+9       =12p+13p+1-p+9       =12p+12p-6
 
The given equation will have real and equal roots if D = 0.

12p+12p-6=0p+1=0 or 2p-6=0p=-1 or p=3

But, p-1           (Given)

Thus, the value of p is 3.

Putting p = 3, the given equation becomes 4x2-24x+36=0.

4x2-24x+36=04x2-6x+9=0x-32=0x-3=0
x=3

Hence, 3 is the repeated root of this equation.

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Question 11:

If −5 is a root of the quadratic equation 2x2+px-15=0 and the quadratic equation px2+x+k=0 has equal roots, find the value of k.
                                                                                                                                                                                                    [CBSE 2014]

Answer:

It is given that −5 is a root of the quadratic equation 2x2+px-15=0.

2-52+p×-5-15=0-5p+35=0p=7

The roots of the equation px2+px+k=0 = 0 are equal.

D=0p2-4pk=072-4×7×k=049-28k=0
k=4928=74

Thus, the value of k is 74.

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Question 12:

If 3 is a root of the quadratic equation x2-x+k=0, find the value of p so that the roots of the equation x2+k2x+k+2+p=0 are equal.
                                                                                                                                                                                            [CBSE 2015]

Answer:

It is given that 3 is a root of the quadratic equation x2-x+k=0.

32-3+k=0k+6=0k=-6

The roots of the equation x2+2kx+k2+2k+p=0 are equal.

D=02k2-4×1×k2+2k+p=04k2-4k2-8k-4p=0-8k-4p=0
p=8k-4=-2kp=-2×-6=12

Hence, the value of p is 12.



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Question 13:

If −4 is a root of the quadratic equation x2+2x+4p=0, find the value of k for which the quadratic equation x2+px1+3k+73+2k=0 has equal roots.                                                                                                                                                                             [CBSE 2015]

Answer:

It is given that −4 is a root of the quadratic equation x2+2x+4p=0.

-42+2×-4+4p=016-8+4p=04p+8=0p=-2

The equation x2+px1+3k+73+2k=0 has equal roots.

 D=0p1+3k2-4×1×73+2k=0-21+3k2-283+2k=041+6k+9k2-283+2k=0
41+6k+9k2-21-14k=09k2-8k-20=09k2-18k+10k-20=09kk-2+10k-2=0
k-29k+10=0k-2=0 or 9k+10=0k=2 or k=-109

Hence, the required value of k is 2 or -109.

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Question 14:

If the equation (1+m2)x2+2mcx+(c2-a2)=0 has equal roots, prove that c2=a2(1+m2).

Answer:

Given: (1 + m2)x2 + 2mcx + (c2  a2) = 0Here, a = (1 + m2), b = 2mc and c = (c2  a2)It is given that the roots of the equation are equal; therefore, we have:D = 0 (b2  4ac) = 0 (2mc)2  4 × (1 + m2) × (c2  a2) = 0 4m2c2  4(c2  a2 + m2c2  m2a2) = 0 4m2c2  4c2 + 4a2  4m2c2 + 4m2a2 = 0 4c2 + 4a2 + 4m2a2 = 0 a2 + m2a2 = c2 a2(1 + m2) = c2 c2 = a2(1 + m2)Hence proved.

Page No 470:

Question 15:

If the roots of the equation (c2-ab)x2-2(a2-bc)x+(b2-ac)=0 are real and equal, show that either a=0 or (a3+b3+c3)=3abc.

Answer:

Given: (c2  ab)x2  2(a2  bc)x + (b2  ac) = 0Here, a = (c2  ab), b = 2(a2  bc), c = (b2  ac)It is given that the roots of the equation are real and equal; therefore, we have:D=0(b2  4ac) = 0 {2(a2  bc)}2  4 × (c2  ab) × (b2  ac) = 0 4(a4  2a2bc + b2c2)  4(b2c2  ac3  ab3 + a2bc) = 0 a4  2a2bc + b2c2  b2c2 + ac3 + ab3  a2bc = 0 a4  3a2bc + ac3 + ab3 = 0 a(a3  3abc + c3 + b3) = 0Now,a = 0 or a3  3abc + c3 + b3 = 0a = 0 or a3 + b3 + c3 = 3abc

Page No 470:

Question 16:

Determine the values of p for which the quadratic equation 2x2+px+8=0 has real roots.

Answer:

Given: 2x2 + px + 8 = 0Here, a = 2, b = p and c = 8Discriminant D is given by:D = (b2  4ac)p2  4 × 2 × 8(p2  64)If D  0, the roots of the equation will be real. (p2  64)  0 (p + 8) (p - 8)  0 p  8 and p  -8Thus, the roots of the equation are real for p  8 and p-8.

Page No 470:

Question 17:

Find the value of α for which the equation (α-12)x2+2(α-12)x+2=0 has equal roots.

Answer:

Given: (α  12)x2 + 2(α  12)x + 2 = 0Here, a = (α  12), b = 2(α  12) and c = 2It is given that the roots of the equation are equal; therefore, we have:D = 0 (b2  4ac) = 0 {2(α  12)}2  4 × (α  12) × 2 = 0 4(α2  24α + 144)  8(α  12) = 0 4α2  96α + 576  8α + 96 = 0 4α2  104α + 672 = 0 α2  26α + 168 = 0 α2  14α  12α + 168 = 0 α(α  14)  12(α  14) = 0 (α  14)(α  12) = 0 α = 14 or α = 12If the value of α is 12, the given equation becomes non-quadratic.Therefore, the valueof α will be 14 for the equation to have equal roots.

Page No 470:

Question 18:

Find the values k for which of roots of 9x2+8kx+16=0 are real and equal

Answer:

Given: 9x2 + 8kx + 16 = 0Here,a = 9, b = 8k and c = 16It is given that the roots of the equation are real and equal; therefore, we have:D = 0 (b2  4ac) = 0 (8k)2  4 × 9 × 16 = 0 64k2  576 = 0 64k2 = 576 k2 = 9 k = ±3 k = 3 or k = 3

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Question 19:

Find the values of k for which the given quadratic equation has real and distinct roots:

(i) kx2+6x+1=0
(ii) x2-kx+9=0
(iii) 9x2+3kx+4=0
(iv) 5x2-kx+1=0

Answer:

(i) The given equation is kx2+6x+1=0.

D=62-4×k×1=36-4k

The given equation has real and distinct roots if D > 0.

36-4k>04k<36k<9

(ii) The given equation is x2-kx+9=0.

D=-k2-4×1×9=k2-36

The given equation has real and distinct roots if D > 0.

k2-36>0k-6k+6>0k<-6 or k>6

(iii) The given equation is 9x2+3kx+4=0.

D=3k2-4×9×4=9k2-144

The given equation has real and distinct roots if D > 0.

9k2-144>09k2-16>0k-4k+4>0k<-4 or k>4

(iv) The given equation is 5x2-kx+1=0.

D=-k2-4×5×1=k2-20

The given equation has real and distinct roots if D > 0.

k2-20>0k2-252>0k-25k+25>0k<-25 or k>25

Page No 470:

Question 20:

If a and b are real and ab then show that the roots of the equation a-bx2+5a+bx-2a-b=0 are real and unequal.

Answer:

The given equation is a-bx2+5a+bx-2a-b=0.

D=5a+b2-4×a-b×-2a-b       =25a+b2+8a-b2
Since a and b are real and ab, so a-b2>0 and a+b2>0.

8a-b2>0    .....(1)               (Product of two positive numbers is always positive)      

Also, 25a+b2>0             .....(2)              (Product of two positive numbers is always positive)

Adding (1) and (2), we get

25a+b2+8a-b2>0                 (Sum of two positive numbers is always positive)

D>0

Hence, the roots of the given equation are real and unequal.

Page No 470:

Question 21:

If the roots of the equation a2+b2x2-2ac+bdx+c2+d2=0 are equal, prove that ab=cd.

Answer:

It is given that the roots of the equation a2+b2x2-2ac+bdx+c2+d2=0 are equal.
D=0-2ac+bd2-4a2+b2c2+d2=04a2c2+b2d2+2abcd-4a2c2+a2d2+b2c2+b2d2=04a2c2+b2d2+2abcd-a2c2-a2d2-b2c2-b2d2=0
-a2d2+2abcd-b2c2=0-a2d2-2abcd+b2c2=0ad-bc2=0
ad-bc=0ad=bcab=cd
Hence Proved.

Page No 470:

Question 22:

If the roots of the equations ax2+2bx+c=0 and bx2-2acx+b=0 are simultaneously real then prove that b2=ac.

Answer:

It is given that the roots of the equation ax2+2bx+c=0 are real.
D1=2b2-4×a×c04b2-ac0b2-ac0             .....1

Also, the roots of the equation bx2-2acx+b=0 are real.
D2=-2ac2-4×b×b04ac-b20-4b2-ac0
b2-ac0                    .....2

The roots of the given equations are simultaneously real if (1) and (2) holds true together. This is possible if
b2-ac=0b2=ac



Page No 491:

Question 1:

The sum of a natural number and its square is 156. Find the number.

Answer:

Let the required natural number be x.

According to the given condition,

x+x2=156x2+x-156=0x2+13x-12x-156=0xx+13-12x+13=0
x+13x-12=0x+13=0 or x-12=0x=-13 or x=12

x = 12     (x cannot be negative)

Hence, the required natural number is 12.

Page No 491:

Question 2:

The sum of a natural number and its positive square root is 132. Find the number.

Answer:

Let the required natural number be x.

According to the given condition,

x+x=132

Putting x=y or x = y2, we get

y2+y=132y2+y-132=0y2+12y-11y-132=0yy+12-11y+12=0
y+12y-11=0y+12=0 or y-11=0y=-12 or y=11

y = 11         (y cannot be negative)

Now,

x=11x=112=121

Hence, the required natural number is 121.

Page No 491:

Question 3:

The sum of two natural numbers is 28 and their product is 192. Find the numbers.

Answer:

Let the required numbers be x and (28 − x).

According to the given condition,

x28-x=19228x-x2=192x2-28x+192=0x2-16x-12x+192=0
xx-16-12x-16=0x-12x-16=0x-12=0 or x-16=0x=12 or x=16

When x = 12,

28 − x = 28 − 12 = 16

When x = 16,

28 − x = 28 − 16 = 12

Hence, the required numbers are 12 and 16.

Page No 491:

Question 4:

The sum of the squares of two consecutive positive integers is 365. Find the integers.

Answer:

Let the required two consecutive positive integers be x and (x + 1).

According to the given condition,
x2+x+12=365x2+x2+2x+1=3652x2+2x-364=0x2+x-182=0
x2+14x-13x-182=0xx+14-13x+14=0x+14x-13=0x+14=0 or x-13=0
x=-14 or x=13
x = 13                (x is a positive integer)

When x = 13,
x + 1 = 13 + 1 = 14

Hence, the required positive integers are 13 and 14.

Page No 491:

Question 5:

The sum of the squares of two consecutive positive odd numbers is 514. Find the numbers.

Answer:

Let the two consecutive positive odd numbers be x and (x + 2).

According to the given condition,
x2+x+22=514x2+x2+4x+4=5142x2+4x-510=0x2+2x-255=0
x2+17x-15x-255=0xx+17-15x+17=0x+17x-15=0x+17=0 or x-15=0
x=-17 or x=15
x = 15             (x is a positive odd number)

When x = 15,
x + 2 = 15 + 2 = 17

Hence, the required numbers are 15 and 17.

Page No 491:

Question 6:

The sum of the squares of two consecutive positive even numbers is 452. Find the numbers.

Answer:

Let the two consecutive positive even numbers be x and (x + 2).

According to the given condition,
x2+x+22=452x2+x2+4x+4=4522x2+4x-448=0x2+2x-224=0
x2+16x-14x-224=0xx+16-14x+16=0x+16x-14=0x+16=0 or x-14=0
x=-16 or x=14
x = 14             (x is a positive even number)

When x = 14,
x + 2 = 14 + 2 = 16

Hence, the required numbers are 14 and 16.

Page No 491:

Question 7:

The product of two consecutive positive integers is 306. Find the integers.

Answer:

Let the two consecutive positive integers be x and (x + 1).

According to the given condition,
xx+1=306x2+x-306=0x2+18x-17x-306=0xx+18-17x+18=0
x+18x-17=0x+18=0 or x-17=0x=-18 or x=17
x = 17           (x is a positive integer)

When x = 17,
x + 1 = 17 + 1 = 18

Hence, the required integers are 17 and 18.

Page No 491:

Question 8:

Two natural numbers differ by 3 and their product is 504. Find the numbers.

Answer:

 Let the required numbers be x and (x+3).According to the question: x(x + 3) = 504 x2 + 3x = 504 x2 + 3x  504 = 0 x2 + (24  21)x  504 = 0 x2 + 24x  21x  504 = 0 x(x + 24)  21(x + 24) = 0 (x + 24)(x  21) = 0 x + 24 = 0 or x  21 = 0 x = 24 or x =  21If x 24, the numbers are 24 and {(24 + 3) 21}.If x = 21, the numbers are 21 and {(21 + 3) = 24}.Hence, the numbers are (24, 21) and (21, 24).

Page No 491:

Question 9:

Find two consecutive multiples of 3 whose product is 648.

Answer:

Let the required consecutive multiples of 3 be 3x and 3(x + 1).

According to the given condition,
3x×3x+1=6489x2+x=648x2+x=72x2+x-72=0
x2+9x-8x-72=0xx+9-8x+9=0x+9x-8=0x+9=0 or x-8=0
x=-9 or x=8
x = 8           (Neglecting the negative value)

When x = 8,
3x = 3 × 8 = 24
3(x + 1) = 3 × (8 + 1) = 3 × 9 = 27

Hence, the required multiples are 24 and 27.

Page No 491:

Question 10:

Find two consecutive positive odd integers whose product is 483.

Answer:

Let the two consecutive positive odd integers be x and (x + 2).

According to the given condition,
xx+2=483x2+2x-483=0x2+23x-21x-483=0xx+23-21x+23=0
x+23x-21=0x+23=0 or x-21=0x=-23 or x=21
x = 21           (x is a positive odd integer)

When x = 21,
x + 2 = 21 + 2 = 23

Hence, the required integers are 21 and 23.

Page No 491:

Question 11:

Find two consecutive positive even integers whose product is 288.

Answer:

Let the two consecutive positive even integers be x and (x + 2).

According to the given condition,
xx+2=288x2+2x-288=0x2+18x-16x-288=0xx+18-16x+18=0
x+18x-16=0x+18=0 or x-16=0x=-18 or x=16
x = 16           (x is a positive even integer)

When x = 16,
x + 2 = 16 + 2 = 18

Hence, the required integers are 16 and 18.

Page No 491:

Question 12:

The sum of two natural numbers is 9 and the sum of their reciprocals is 12. Find the numbers.                              [CBSE 2012]

Answer:

Let the required natural numbers be x and (9 − x).

According to the given condition,
1x+19-x=129-x+xx9-x=1299x-x2=129x-x2=18
x2-9x+18=0x2-6x-3x+18=0xx-6-3x-6=0x-3x-6=0
x-3=0 or x-6=0x=3 or x=6

When x = 3,
9 − x = 9 − 3 = 6

When x = 6,
9 − x = 9 − 6 = 3

Hence, the required natural numbers are 3 and 6.

Page No 491:

Question 13:

The sum of two natural numbers is 15 and the sum of their reciprocals is 310. Find the numbers.                              [CBSE 2005]

Answer:

Let the required natural numbers be x and (15 − x).

According to the given condition,
1x+115-x=31015-x+xx15-x=3101515x-x2=31015x-x2=50
x2-15x+50=0x2-10x-5x+50=0xx-10-5x-10=0x-5x-10=0
x-5=0 or x-10=0x=5 or x=10

When x = 5,
15 − x = 15 − 5 = 10

When x = 10,
15 − x = 15 − 10 = 5

Hence, the required natural numbers are 5 and 10.

Page No 491:

Question 14:

The difference of two natural numbers is 3 and the difference of their reciprocals is 328. Find the numbers.                            [CBSE 2014]

Answer:

Let the required natural numbers be x and (x + 3).

Now, x < x + 3
1x>1x+3

According to the given condition,
1x-1x+3=328x+3-xxx+3=3283x2+3x=328x2+3x=28
x2+3x-28=0x2+7x-4x-28=0xx+7-4x+7=0x+7x-4=0
x+7=0 or x-4=0x=-7 or x=4
x = 4             (−7 is not a natural number)

When x = 4,
x + 3 = 4 + 3 = 7

Hence, the required natural numbers are 4 and 7.

Page No 491:

Question 15:

The difference of two natural numbers is 5 and the difference of their reciprocals is 514. Find the numbers.                            [CBSE 2014]

Answer:

Let the required natural numbers be x and (x + 5).

Now, x < x + 5
1x>1x+5

According to the given condition,
1x-1x+5=514x+5-xxx+5=5145x2+5x=514x2+5x=14
x2+5x-14=0x2+7x-2x-14=0xx+7-2x+7=0x+7x-2=0
x+7=0 or x-2=0x=-7 or x=2
x = 2             (−7 is not a natural number)

When x = 2,
x + 5 = 2 + 5 = 7

Hence, the required natural numbers are 2 and 7.

Page No 491:

Question 16:

The sum of the squares of two consecutive multiples of 7 is 1225. Find the multiples.

Answer:

Let the required consecutive multiples of 7 be 7x and 7(x + 1).

According to the given condition,
7x2+7x+12=122549x2+49x2+2x+1=122549x2+49x2+98x+49=122598x2+98x-1176=0
x2+x-12=0x2+4x-3x-12=0xx+4-3x+4=0x+4x-3=0
x+4=0 or x-3=0x=-4 or x=3
x = 3            (Neglecting the negative value)

When x = 3,
7x = 7 × 3 = 21
7(x + 1) = 7(3 + 1) = 7 × 4 = 28

Hence, the required multiples are 21 and 28.

Page No 491:

Question 17:

The sum of a natural number and its reciprocal is 658. Find the number.

Answer:

Let the natural number be x.

According to the given condition,
x+1x=658x2+1x=6588x2+8=65x8x2-65x+8=0
8x2-64x-x+8=08xx-8-1x-8=0x-88x-1=0x-8=0 or 8x-1=0
x=8 or x=18
x = 8         (x is a natural number)

Hence, the required number is 8.

Page No 491:

Question 18:

Divide 57 into two parts whose product is 680.

Answer:

Let the two parts be x and (57 − x).

According to the given condition,
x57-x=68057x-x2=680x2-57x+680=0x2-40x-17x+680=0
xx-40-17x-40=0x-40x-17=0x-40=0 or x-17=0x=40 or x=17

When x = 40,
57 − x = 57 − 40 = 17

When x = 17,
57 − x = 57 − 17 = 40

Hence, the required parts are 17 and 40.



Page No 492:

Question 19:

Divide 27 into two parts such that the sum of their reciprocals is 320.

Answer:

Let the two parts be x and (27 − x).

According to the given condition,
1x+127-x=32027-x+xx27-x=3202727x-x2=32027x-x2=180
x2-27x+180=0x2-15x-12x+180=0xx-15-12x-15=0x-12x-15=0
x-12=0 or x-15=0x=12 or x=15

When x = 12,
27 − x = 27 − 12 = 15

When x = 15,
27 − x = 27 − 15 = 12

Hence, the required parts are 12 and 15.

Page No 492:

Question 20:

Divide 16 into two parts such that twice the squares of the larger part exceeds the square of the smaller part by 164.

Answer:

Let the larger and smaller parts be x and y, respectively.According to the question:x y = 16 ...(i)2x2 = y2 + 164 ...(ii)From (i), we get: x = 16  y ...(iii) From (ii) and (iii), we get:2(16  y)2 = y2 + 164 2(256  32y + y2) = y2 + 164 512  64y + 2y2 = y2 + 164 y2  64y + 348 = 0 y2  (58 + 6)y + 348 = 0 y2  58y  6y + 348 = 0  y(y  58)  6(y  58) = 0 (y  58)(y  6) = 0 y  58 = 0 or y  6 = 0 y = 6   ( y < 16)Putting the value of y in equation (iii), we get:  x = 16  6 = 10Hence, the two natural numbers are 6 and 10.

Page No 492:

Question 21:

Find two natural numbers, the sum of whose squares is 25 times their sum and also equal to 50 times their difference.

Answer:

 Let the two natural numbers be x and  y.According to the question:x2 + y2 = 25(x + y) ...(i)x2 + y2 = 50(x  y)   ...(ii)From (i) and (ii), we get:25(x + y) = 50(x  y) x + y = 2(x  y) x + y = 2x  2y y + 2y = 2x  x 3y = x     ...(iii)  From (ii) and (iii), we get:(3y)2 + y2 = 50(3y  y) 9y2 + y2 = 100y 10y2 = 100y y = 10 From (iii), we have:3 × 10 = x 30 = x Hence, the two natural numbers are 30 and 10.

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Question 22:

The difference of the squares of two natural numbers is 45. The squares of the smaller number is four times the largest number. Find the numbers.

Answer:

Let the greater number be x and the smaller number be y.According to the question:x2  y2 = 45          ...(i)y2 = 4x                    ...(ii)From (i) and (ii), we get:x2  4x = 45 x2  4x  45 = 0 x2  (9  5)x  45 = 0 x2  9x + 5x  45 = 0 x(x  9) + 5(x  9) = 0 (x  9)(x + 5) = 0 x  9 = 0 or x + 5 = 0 x = 9 or x = 5 x = 9      ( x is a natural number)Putting the value of x in equation (ii), we get:y2 = 4 × 9 y2 = 36 y = 6Hence, the two numbers are 9 and 6.

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Question 23:

Three consecutive positive integers are such that the sum of the square of the first and the product of the other two is 46. Find the integers.                                                                                                                                                                                       [CBSE 2010]

Answer:

Let the three consecutive positive integers be x, x + 1 and x + 2.

According to the given condition,
x2+x+1x+2=46x2+x2+3x+2=462x2+3x-44=02x2+11x-8x-44=0
x2x+11-42x+11=02x+11x-4=02x+11=0 or x-4=0x=-112 or x=4
x = 4           (x is a positive integer)

When x = 4,
x + 1 = 4 + 1 = 5
x + 2 = 4 + 2 = 6

Hence, the required integers are 4, 5 and 6.

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Question 24:

A two-digit number is 4 times the sum of its digits and twice the product of its digit. Find the number.

Answer:

 Let the digits at units and tens places be x and y, respectively. Original number = 10y + x According to the question:10y + x = 4(x + y) 10y + x = 4x + 4y 3x  6y = 0 3x = 6y x = 2y ....(i) Also,10y x = 2xy 10y + 2y = 2.2y.y           [ From (i)] 12y = 4y2 y = 3 From (i), we get:x = 2 × 3 =  6 Original number = 10 × 3 + 6 = 36

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Question 25:

A two-digit number is such that the product of its digits is 14. If 45 is added to the number, the digits interchange their places. Find the number.

Answer:

Let the digits at units and tens places be x and y, respectively. ∴ xy = 14 ⇒ y 14x              ...(i) According to the question:(10y + x) + 45 = 10x + y 9y  9x =  45 y  x =  5                                      ...(ii) From (i) and (ii), we get:14x  x =  5 14  x2x =  5 14  x2 =  5x x2  5x  14 = 0 x2  (7  2)x  14 = 0 x2  7x + 2x  14 = 0 x(x  7) + 2(x  7) = 0 (x  7)(x + 2) = 0 x  7 = 0 or x + 2 = 0 x = 7 or x =  2 x = 7   ( the digit cannot be negative) Putting x = 7 in equation (i), we get:y = 2∴ Required number = 10 × 2 + 7 = 27

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Question 26:

The denominator of a fraction is 3 more than its numerator. The sum of the fraction and its reciprocal is 2910. Find the fraction.

Answer:

 Let the numerator be x. Denominator = x + 3 Original number = xx + 3 According to the question:xx + 3 +  1(xx + 3) =  2910 xx + 3 + x + 3x = 2910 x2 + (x + 3)2x(x + 3) = 2910 x2 + x2 + 6x + 9x2 + 3x = 2910 2x2 + 6x + 9x2 + 3x = 2910 29x2 + 87x = 20x2 + 60x + 90 9x2 + 27x  90 = 0 9x2 + 3x - 10 = 0x2 + 3x - 10 = 0x2 + 5x - 2x - 10 = 0xx+5-2x+5 = 0x-2x+5 = 0x-2 = 0  or  x + 5 = 0x=2 or x = -5rejectedSo, numerator = x = 2denominator = x + 3 = 2 + 3 = 5So, required fraction = 25

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Question 27:

The numerator of a fraction is 3 less than its denominator. If 1 is added to the denominator, the fraction is decreased by 115. Find the fraction.                                                                                                                                                                                          [CBSE 2012]

Answer:

Let the denominator of the required fraction be x.

Numerator of the required fraction = x − 3

∴ Original fraction = x-3x
If 1 is added to the denominator, then the new fraction obtained is x-3x+1.

According to the given condition,
x-3x+1=x-3x-115x-3x-x-3x+1=115x-3x+1-xx-3xx+1=115x2-2x-3-x2+3xx2+x=115
x-3x2+x=115x2+x=15x-45x2-14x+45=0x2-9x-5x+45=0
xx-9-5x-9=0x-5x-9=0x-5=0 or x-9=0x=5 or x=9

When x = 5,
x-3x=5-35=25

When x = 9,
x-3x=9-39=69=23             (This fraction is neglected because this does not satisfies the given condition.)

Hence, the required fraction is 25.

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Question 28:

The sum of a number and its reciprocal is 2130. Find the number.

Answer:

Let the required number be x.

According to the given condition,
x+1x=2130x2+1x=613030x2+30=61x30x2-61x+30=0
30x2-36x-25x+30=06x5x-6-55x-6=05x-66x-5=05x-6=0 or 6x-5=0
x=65 or x=56
Hence, the required number is 56 or 65.

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Question 29:

A teacher on attempting to arrange the students for mass drill in the form of a solid square found that 24 students were left. When he increased the size of the square by one student, he found that he was short of 25 students. Find the number of students.

Answer:

Let there be x rows.Then, the number of students in each row will also be x. Total number of students = (x2 + 24)According to the question:(x + 1)2  25 = x2 + 24 x2 + 2x + 1  25  x2  24 =  0 2x  48 = 0 2x = 48 x = 24 Total number of students = 242 + 24 = 576 + 24 = 600

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Question 30:

300 apples are distributed equally among a certain number of students. Had there been 10 more students, each would have received one apple less. Find the number of student.

Answer:

Let the total number of students be x.According to the question:300x  300x + 10 = 1300(x + 10)  300xx(x + 10) = 1 300x + 3000  300xx2 + 10x = 1 3000 = x2 + 10x x2 + 10x  3000 = 0 x2 + (60  50)x  3000 = 0 x2 + 60x  50x  3000 = 0 x(x + 60)  50(x + 60) = 0 (x + 60)(x  50) = 0 x = 50 or x =  60x cannot be negative; therefore, the total number of students is 50.

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Question 31:

In a class test, the sum of Kamal's marks in Mathematics and English is 40. Had he got 3 marks more in Mathematics and 4 marks less in English, the product of the marks would have been 360. Find his marks in two subjects separately.

Answer:

Let the marks of Kamal in mathematics and english be x and y, respectively.According to the question:x + y = 40         ...(i) Also,(x + 3)(y  4) = 360 (x + 3)(40  x  4) = 360    [From(i)] (x + 3)(36  x) = 360 36x  x2 + 108  3x = 360 33x  x2  252 = 0 x2 + 33x  252 = 0 x2  33x + 252 = 0 x2  (21 + 12)x + 252 = 0 x2  21x  12x + 252 = 0 x(x  21)  12(x  21) = 0 (x  21)(x  12) = 0 x = 21 or x = 12If x = 21, y = 40  21 = 19Thus, Kamal scored 21 and 19 marks in mathematics and english, respectively.If x = 12, y = 40  12 = 28Thus, Kamal scored 12 and 28 marks in mathematics and english, respectively.



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Question 32:

Some students planned a picnic. The total budget for food was ₹2000. But, 5 students failed to attend the picnic and thus the cost for food for each member increased by ₹20. How many students attended the picnic and how much did each student pay for the food?                                                                                                                                                                                                    [CBSE 2010]

Answer:

Let x be the number of students who planned a picnic.

∴ Original cost of food for each member = ₹2000x

Five students failed to attend the picnic. So, (x − 5) students attended the picnic.

∴ New cost of food for each member = ₹2000x-5

According to the given condition,
2000x-5 − ₹2000x = ₹20
2000x-2000x+10000xx-5=2010000x2-5x=20x2-5x=500x2-5x-500=0
x2-25x+20x-500=0xx-25+20x-25=0x-25x+20=0x-25=0 or x+20=0
x=25 or x=-20
x = 25                (Number of students cannot be negative)

Number of students who attended the picnic = x − 5 = 25 − 5 = 20

Amount paid by each student for the food = ₹200025-5 = ₹200020 = ₹100

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Question 33:

If the price of a book is reduced by ₹5, a person can buy 4 more books for ₹600. Find the original price of the book.

Answer:

Let the original price of the book be ₹x.

∴ Number of books bought at original price for ₹600 = 600x

If the price of a book is reduced by ₹5, then the new price of the book is ₹(x − 5).

∴ Number of books bought at reduced price for ₹600 = 600x-5

According to the given condition,
600x-5-600x=4600x-600x+3000xx-5=43000x2-5x=4x2-5x=750
x2-5x-750=0x2-30x+25x-750=0xx-30+25x-30=0x-30x+25=0
x-30=0 or x+25=0x=30 or x=-25
x = 30               (Price cannot be negative)

Hence, the original price of the book is ₹30.

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Question 34:

A person on tour has ₹10800 for his expenses. If he extends his tour by 4 days, he has to cut down his daily expenses by ₹90. Find the original duration of the tour.

Answer:

Let the original duration of the tour be x days.

∴ Original daily expenses = ₹10,800x

If he extends his tour by 4 days, then his new daily expenses = ₹10,800x+4

According to the given condition,

10,800x − ₹10,800x+4 = ₹90

10800x+43200-10800xxx+4=9043200x2+4x=90x2+4x=480x2+4x-480=0
x2+24x-20x-480=0xx+24-20x+24=0x+24x-20=0x+24=0 or x-20=0
x=-24 or x=20
x = 20              (Number of days cannot be negative)

Hence, the original duration of the tour is 20 days.

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Question 35:

In a class test, the sum of the marks obtained by P in mathematics and science is 28. Had he got 3 more marks in mathematics and 4 marks less in science, the product of marks obtained in the two subjects would have been 180. Find the marks obtained him in the two subjects separately.                                                                                                                                                                                           [CBSE 2008]

Answer:

Let the marks obtained by P in mathematics and science be x and (28 − x), respectively.

According to the given condition,
x+328-x-4=180x+324-x=180-x2+21x+72=180x2-21x+108=0
x2-12x-9x+108=0xx-12-9x-12=0x-12x-9=0x-12=0 or x-9=0
x=12 or x=9

When x = 12,
28 − x = 28 − 12 = 16

When x = 9,
28 − x = 28 − 9 = 19

Hence, he obtained 12 marks in mathematics and 16 marks in science or 9 marks in mathematics and 19 marks in science.

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Question 36:

A man busy a number of pens for Rs 80. If he had bought 4 more pens for the same amount, each pen would have cost him Rs 1 less. How many pens did he buy?

Answer:

Let the total number of pens be x.According to the question:80x  80x + 4 = 1 80(x + 4)  80xx(x + 4) = 1 80x + 320  80xx2 + 4x = 1 320 = x2 + 4x x2 + 4x  320 = 0 x2 + (20  16)x  320 = 0 x2 + 20x  16x  320 = 0 x(x + 20)  16(x + 20) = 0 (x + 20)(x  16) = 0 x = 20 or x = 16The total number of pens cannot be negative; therefore, the total number of pens is 16.

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Question 37:

A dealer sells an article for ₹75 and gains as much percent as the cost price of the article. Find the cost price of the article.           
                                                                                                                                                                                                         [CBSE 2011]

Answer:

Let the cost price of the article be ₹x.

∴ Gain percent = x%

According to the given condition,
x + ₹x100×x = ₹75                            (Cost price + Gain = Selling price)
100x+x2100=75x2+100x=7500x2+100x-7500=0x2+150x-50x-7500=0
xx+150-50x+150=0x-50x+150=0x-50=0 or x+150=0x=50 or x=-150
x = 50                   (Cost price cannot be negative)

Hence, the cost price of the article is ₹50.

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Question 38:

One year ago, a man was 8 times as old as his son. Now, his age is equal to the square of his son's age. Find their present ages.

Answer:

Let the present age of the son be x years.

∴ Present age of the man = x2 years

One year ago,

Age of the son = (x − 1) years

Age of the man = (x2 − 1) years

According to the given condition,

Age of the man = 8 × Age of the son

x2-1=8x-1x2-1=8x-8x2-8x+7=0x2-7x-x+7=0
xx-7-1x-7=0x-1x-7=0x-1=0 or x-7=0x=1 or x=7
x = 7                (Man's age cannot be 1 year) 

Present age of the son = 7 years

Present age of the man = 72 years = 49 years

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Question 39:

The sum of the reciprocals of Meena's ages (in years) 3 years ago and 5 years hence is 13. Find her present age.

Answer:

Let the present age of Meena be x years.

Meena's age 3 years ago = (x − 3) years

Meena's age 5 years hence = (x + 5) years

According to the given condition,
1x-3+1x+5=13x+5+x-3x-3x+5=132x+2x2+2x-15=13x2+2x-15=6x+6
x2-4x-21=0x2-7x+3x-21=0xx-7+3x-7=0x-7x+3=0
x-7=0 or x+3=0x=7 or x=-3
x = 7                 (Age cannot be negative)

Hence, the present age of Meena is 7 years.

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Question 40:

The sum of the ages of a boy and his brother is 25 years, and the product of their ages in years is 126. Find their ages.

Answer:

Let the present ages of the boy and his brother be x years and (25  x) years.According to the question:x(25  x) = 126 25x  x2 = 126 x2  (18 + 7)x + 126 = 0 x2  18x  7x + 126 = 0 x(x  18)  7(x  18) = 0 (x  18)(x  7) = 0 x  18 = 0 or  x  7 = 0 x = 18 or x = 7 x = 18        ( Present age of the boy cannot be less than his brother)If x=18, we have:Present ages of the boy = 18 yearsPresent age of his brother = (2518) years = 7 yearsThus, the present ages of the boy and his brother are 18 years and 7 years, respectively.

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Question 41:

The product of Tanvy's age (in years) 5 years ago and her age 8 years later is 30. Find her present age.

Answer:

Let the present age of Meena be x years. According to the question:(x  5)(x + 8) = 30 x2 + 3x  40 = 30 x2 + 3x  70 = 0 x2 + (10  7)x  70 = 0 x2 + 10x  7x  70 = 0 x(x + 10)  7(x + 10) = 0 (x + 10)(x  7) = 0 x + 10 = 0 or  x  7 = 0 x = 10 or x = 7 x = 7      (  Age cannot be negative)Thus, the present age of Meena is 7 years.

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Question 42:

Two years ago, a man's age was three times the square of his son's age. In three years time, his age will be four times his son's age. Find their present ages.

Answer:

Let son's age 2 years ago be x years. Then,

Man's age 2 years ago = 3x2 years

∴ Son's present age = (x + 2) years

Man's present age = (3x2 + 2) years

In three years time,

Son's age = (x + 2 + 3) years = (x + 5) years

Man's age = (3x2 + 2 + 3) years = (3x2 + 5) years

According to the given condition,

Man's age = 4 × Son's age

∴ 3x2 + 5 = 4(x + 5)

3x2+5=4x+203x2-4x-15=03x2-9x+5x-15=03xx-3+5x-3=0
x-33x+5=0x-3=0 or 3x+5=0x=3 or x=-53
x = 3              (Age cannot be negative)

Son's present age = (x + 2) years = (3 + 2) years = 5 years

Man's present age = (3x2 + 2) years = (3 × 9 + 2) years = 29 years

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Question 43:

A truck covers a distance of 150 km at a certain average speed and then covers another 200 km at an average speed which is 20 km per hour more than the first speed. If the truck covers the total distance in 5 hours, find the first speed of the truck.                                 [CBSE 2015]

Answer:

Let the first speed of the truck be x km/h.

∴ Time taken to cover 150 km = 150x h                                 Time =DistanceSpeed

New speed of the truck = (x + 20) km/h

∴ Time taken to cover 200 km = 200x+20 h

According to the given condition,

Time taken to cover 150 km + Time taken to cover 200 km = 5 h
150x+200x+20=5150x+3000+200xxx+20=5350x+3000=5x2+20x350x+3000=5x2+100x
5x2-250x-3000=0x2-50x-600=0x2-60x+10x-600=0xx-60+10x-60=0
x-60x+10=0x-60=0 or x+10=0x=60 or x=-10
x = 60                  (Speed cannot be negative)

Hence, the first speed of the truck is 60 km/h.



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Question 44:

While boarding an aeroplane, a passenger got hurt. The pilot showing promptness and concern, made arrangements to hospitalise the injured and so the plane started late by 30 minutes. To reach the destination, 1500 km away, in time, the pilot increased the speed by 100 km/hour.
Find the original speed of the plane.
Do you appreciate the values shown by the pilot, namely promptness in providing help to the injured and his efforts to reach in time?   
                                                                                                                                                                                                           [CBSE 2013]

Answer:

Let the original speed of the plane be x km/h.

∴ Actual speed of the plane = (x + 100) km/h

Distance of the journey = 1500 km

Time taken to reach the destination at original speed = 1500x h                             Time=DistanceSpeed

Time taken to reach the destination at actual speed = 1500x+100 h

According to the given condition,

Time taken to reach the destination at original speed = Time taken to reach the destination at actual speed + 30 min

1500x=1500x+100+12                          30 min=3060h=12h1500x-1500x+100=121500x+150000-1500xxx+100=12150000x2+100x=12
x2+100x=300000x2+100x-300000=0x2+600x-500x-300000=0xx+600-500x+600=0
x+600x-500=0x+600=0 or x-500=0x=-600 or x=500
x = 500                 (Speed cannot be negative)

Hence, the original speed of the plane is 500 km/h.

Yes, we appreciate the values shown by the pilot, namely promptness in providing help to the injured and his efforts to reach in time. This reflects the caring nature of the pilot and his dedication to the work.

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Question 45:

A train covers a distance of 480 km at a uniform speed. If the speed had been 8 km/h less then it would have taken 3 hours more to cover the same distance. Find the usual speed of the train.                                                                                                                             [CBSE 2013C]

Answer:

Let the usual speed of the train be x km/h.

∴ Reduced speed of the train = (x − 8) km/h

Total distance to be covered = 480 km

Time taken by the train to cover the distance at usual speed = 480x h                  Time=DistanceSpeed

Time taken by the train to cover the distance at reduced speed = 480x-8 h

According to the given condition,

Time taken by the train to cover the distance at reduced speed = Time taken by the train to cover the distance at usual speed + 3 h

480x-8=480x+3480x-8-480x=3480x-480x+3840xx-8=33840x2-8x=3
x2-8x=1280x2-8x-1280=0x2-40x+32x-1280=0xx-40+32x-40=0
x-40x+32=0x-40=0 or x+32=0x=40 or x=-32
x = 40              (Speed cannot be negative)

Hence, the usual speed of the train is 40 km/h.

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Question 46:

A train travels at a certain average speed for a distance of 54 km and then travels a distance of 63 km at an average of 6 km/hr more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed?                                                                        [CBSE 2015]

Answer:

Let the first speed of the train be x km/h.

∴ Time taken to cover 54 km = 54x h                                 Time =DistanceSpeed

New speed of the train = (x + 6) km/h

∴ Time taken to cover 63 km = 63x+6 h

According to the given condition,

Time taken to cover 54 km + Time taken to cover 63 km = 3 h
54x+63x+6=354x+324+63xxx+6=3117x+324=3x2+6x117x+324=3x2+18x
3x2-99x-324=0x2-33x-108=0x2-36x+3x-108=0xx-36+3x-36=0
x-36x+3=0x-36=0 or x+3=0x=36 or x=-3
x = 36                  (Speed cannot be negative)

Hence, the first speed of the train is 36 km/h.

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Question 47:

A train travels 180 km at a uniform speed. If the speed had been 9 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.       [CBSE 2013]

Answer:

Distance covered by the train = 180 km
We know that distance covered(d)=speed(s)×time(t)
s×t=180s=180t          ...i
Also, given that if the speed is increased by 9km/h, time of travel gets reduced by 1 hour. 
(s+9)(t-1)=180          ...(ii)
Put the value of (i) in (ii), we get
180t+9(t-1)=180(180+9t)(t-1)=180t180t-180+9t2-9t=180t9t2-9t-180=0t2-t-20=0t2-5t+4t-20=0t(t-5)+4(t-5)=0(t+4)(t-5)=0(t+4)=0 or (t-5)=0t=-4 or t=5
Ignore the negative value
So, time taken = 5 hours
From (i)
s=180t=1805=36
Hence, the speed is 36 km/h.

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Question 48:

A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hr more, it would have taken 30 minutes less for the journey. Find the original speed of the train.

Answer:

Let the original speed of the train be x km/hr.According to the question:90x  90(x + 15) = 12 90(x + 15)  90xx(x + 15) = 12 90x + 1350  90xx2 + 15x = 12 1350x2 + 15x = 12 2700 = x2 + 15x x2 + (60  45)x  2700 = 0 x2 + 60x  45x  2700 = 0 x(x + 60)  45(x + 60) = 0 (x + 60)(x  45) = 0 x =  60 or x = 45 x cannot be negative; therefore, the original speed of train is 45 km/hr.

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Question 49:

A passenger train takes 2 hours less for a journey of 300 km if its speed is increased by 5 km/hr from its usual speed. Find its usual speed.

Answer:

Let the usual speed be x km/hr.According to the question:300x  300(x + 5) = 2 300(x + 5)  300xx(x + 5) = 2 300x + 1500  300xx2 + 5x = 2 1500 = 2(x2 + 5x) 1500 = 2x2 + 10x x2 + 5x  750 = 0 x2 + (30  25)x  750 = 0 x2 + 30x  25x  750 = 0 x(x + 30)  25(x + 30) = 0 (x + 30)(x  25) = 0 x = 30 or x = 25The usual speed cannot be negative; therefore, the speed is 25 km/hr.

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Question 50:

The distance between Mumbai and Pune is 192 km. Travelling by the Deccan Queen, it takes 48 minutes less than another train. Calculate the speed of the Deccan Queen if the speed of the two trains differ by 20 km/hr.

Answer:

Let the speed of the Deccan Queen be x km/hrAccording to the question:Speed of another train = (x - 20) km/h192x  20  192x = 4860 4x  20  4x = 160 4x  4(x  20)(x  20)x = 160 4x  4x + 80x2  20x = 160 80x2  20x = 160 x2  20x = 4800 x2  20x  4800 = 0 x2  (80  60)x  4800 = 0 x2  80x + 60x  4800 = 0 x(x  80) + 60(x  80) = 0 (x  80)(x + 60) =  0 x = 80 or x = 60 The value of x cannot be negative; therefore, the original speed of Deccan Queen is 80 km/hr.

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Question 51:

A motor boat whose speed in still water is 18 km/hr, takes 1 hour more to go 24 km upstream than o return to the same spot. Find the speed of the stream.

Answer:

Let the speed of the stream be x km/hr.Given:Speed of the boat = 18 km/hr Speed downstream = (18 + x) km/hr     Speed upstream = (18  x) km/hr 24(18  x)  24(18 + x) = 1 1(18  x)  1(18 + x) = 124 18 + x  18 + x(18  x)(18 + x) = 124 2x182  x2 = 124 324  x2 = 48x 324  x2  48x = 0 x2 + 48x  324 = 0 x2 + (54  6)x  324 = 0 x2 + 54x  6x  324 = 0 x(x + 54)  6(x + 54) = 0 (x + 54)(x  6) = 0 x = 54 or x = 6The value of x cannot be negative; thereore, the speed of the stream is 6 km/hr.

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Question 52:

The speed of a boat in still water is 8 km/hr. It can go 15 km upstream and 22 km downstream in 5 hours. Find the speed of the stream.

Answer:

Speed of the boat in still water = 8 km/hrLet the speed of the stream be x km/hr.∴  Speed upstream = (8  x) km/hr     Speed downstream = (8 + x) km/hrTime taken to go 22 km downstream = 22(8 + x) hrTime taken to go 15 km upstream = 15(8  x) hrAccording to the question: 22(8 + x) + 15(8  x) = 5 22(8 + x) + 15(8  x)  5 = 0 22(8  x) + 15(8 + x)  5(8  x)(8 + x)(8  x)(8 + x) = 0 176  22x + 120 + 15x  320 + 5x2 = 0 5x2  7x  24 = 0 5x2  (15  8)x  24 = 0 5x2  15x + 8x  24 = 0 5x(x  3)  8(x  3) = 0 (x  3)(5x  8) = 0 x  3 = 0 or 5x  8 = 0 x = 3 or x 85 x = 3      ( Speed cannot be a fraction) Speed of the stream = 3 km/hr

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Question 53:

A motorboat whose speed is 9 km/hr in still water, goes 15 km downstream and comes back in a total time of 3 hours 45 minutes. Find the speed of the stream.

Answer:

Let the speed of the stream be x km/hr. Downstream speed= (9 + x) km/hrUpstream speed= (9  x) km/hrDistance covered downstream = Distance covered upstream= 15 kmTotal time taken = 3 hours 45 minutes =  3+4560 minutes = 22560 minutes = 154minutes15(9 + x) + 15(9  x) = 154 1(9 + x) + 1(9  x) = 14 9  x + 9 + x(9 + x)(9  x) = 14 1892  x2 = 14 1881  x2 = 14 81  x2 = 72 81  x2  72 = 0 x2 + 9 = 0 x2 = 9 x = 3 or x = 3The value of x cannot be negative; therefore, the speed of the stream is 3 km/hr.



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Question 54:

A takes 10 days than the time taken by B to finish a piece of work. If both A and B together can finish the work in 12 days, find the time taken by B to finish the work.

Answer:

Let B takes x days to complete the work. Therefore, A will take (x  10) days. 1x + 1(x  10) = 112 (x  10) + xx(x  10) = 112 2x  10x2  10x = 112 x2  10x = 12(2x  10) x2  10x = 24x  120 x2  34x + 120 = 0 x2  (30 + 4)x + 120 = 0 x2  30x  4x + 120 = 0 x(x  30)  4(x  30) = 0 (x  30)(x  4) = 0 x = 30 or x = 4Number of days to complete the work by B cannot be less than that by A; therefore, we get:x = 30Thus, B completes the work in 30 days.

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Question 55:

Two pipes running together can fill a cistern in 3113 minutes. If one pipe takes 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern.

Answer:

Let one pipe fills the cistern in x mins.Therefore, the other pipe will fill the cistern in (x + 3) mins.Time taken by both, running together, to fill the cistern = 3113 mins = 4013 minsPart filled by one pipe in 1 min = 1xPart filled by the other pipe in 1 min = 1x + 3Part filled by both pipes, running together, in 1 min = 1x + 1x + 3 1x + 1x + 3 = 14013 (x + 3) + xx(x + 3) = 1340 2x + 3x2 + 3x = 1340 13x2 + 39x = 80x + 120 13x2  41x  120 = 0 13x2  (65  24)x  120 = 0 13x2  65x + 24x  120 = 0 13x(x  5) + 24(x  5) = 0 (x  5)(13x + 24) = 0 x  5 = 0 or 13x + 24 = 0⇒ x = 5 or x 2413 x = 5    ( Speed cannot be a negative fraction)Thus, one pipe will take 5 mins and the other will take {(5 + 3) = 8} mins to fill the cistern.

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Question 56:

Two pipes running together can fill a tank in 1119 minutes. If one pipe takes 5 minutes more than the other to fill the tank separately, find the time in which each pipe would fill the tank separately.                                                                                                                [CBSE 2010]

Answer:

Let the time taken by one pipe to fill the tank be x minutes.

∴ Time taken by the other pipe to fill the tank = (x + 5) min

Suppose the volume of the tank be V.

Volume of the tank filled by one pipe in x minutes = V

∴ Volume of the tank filled by one pipe in 1 minute = Vx

⇒ Volume of the tank filled by one pipe in 1119 minutes = Vx×1119=Vx×1009

Similarly,

Volume of the tank filled by the other pipe in 1119 minutes = Vx+5×1119=Vx+5×1009

Now,

Volume of the tank filled by one pipe in 1119 minutes + Volume of the tank filled by the other pipe in 1119 minutes = V
V1x+1x+5×1009=V1x+1x+5=9100x+5+xxx+5=91002x+5x2+5x=9100
200x+500=9x2+45x9x2-155x-500=09x2-180x+25x-500=09xx-20+25x-20=0
x-209x+25=0x-20=0 or 9x+25=0x=20 or x=-259

x = 20                    (Time cannot be negative)

Time taken by one pipe to fill the tank = 20 min

Time taken by other pipe to fill the tank = (20 + 5) = 25 min

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Question 57:

Two water taps together can fill a tank in 6 hours. The tap of larger diameter takes 9 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.                                                                                                             [CBSE 2011]

Answer:

Let the tap of smaller diameter fill the tank in x hours.

∴ Time taken by the tap of larger diameter to fill the tank = (x − 9) h

Suppose the volume of the tank be V.

Volume of the tank filled by the tap of smaller diameter in x hours = V

∴ Volume of the tank filled by the tap of smaller diameter in 1 hour = Vx

⇒ Volume of the tank filled by the tap of smaller diameter in 6 hours = Vx×6

Similarly,

Volume of the tank filled by the tap of larger diameter in 6 hours = Vx-9×6

Now,

Volume of the tank filled by the tap of smaller diameter in 6 hours + Volume of the tank filled by the tap of larger diameter in 6 hours = V
V1x+1x-9×6=V1x+1x-9=16x-9+xxx-9=162x-9x2-9x=16
12x-54=x2-9xx2-21x+54=0x2-18x-3x+54=0xx-18-3x-18=0
x-18x-3=0x-18=0 or x-3=0x=18 or x=3


For x = 3, time taken by the tap of larger diameter to fill the tank is negative which is not possibe.

x = 18                   

Time taken by the tap of smaller diameter to fill the tank = 18 h

Time taken by the tap of larger diameter to fill the tank = (18 − 9) = 9 h

Hence, the time taken by the taps of smaller and larger diameter to fill the tank is 18 hours and 9 hours, respectively.

Page No 495:

Question 58:

The length of a rectangle is twice its breadth and its area is 288 cm2. Find the dimensions of the rectangle.

Answer:

Let the length and breadth of the rectangle be 2x m and x m, respectively.According to the question:2x × x = 288 2x2 = 288 x2 = 144 x = 12 or x 12 x = 12        ( x cannot be negative) Length = 2 × 12 = 24 m     Breadth = 12 m

Page No 495:

Question 59:

The length of a rectangular field is three times its breadth. If the area of the field by 147 sq metres, find the length of the field.

Answer:

Let the length and breadth of the rectangle be 3x m and x m, respectively.According to the question:3x × x = 147 3x2 = 147 x2 = 49 x = 7 or x 7 x = 7     (x cannot be negative) Length = 3 × 7 = 21m      Breadth = 7 m

Page No 495:

Question 60:

The length of a hall is 3 metres more than its breadth. If the area of the hall is 238 square metres, calculate its length and breadth.

Answer:

Let the breadth of the rectangular hall be x metre.Therefore, the length of the rectangular hall will be (x + 3)metre.According to the question:x(x + 3) = 238 x2 + 3x = 238 x2 + 3x  238 = 0 x2 + (17  14)x  238 = 0 x2 + 17x  14x  238 = 0 x(x + 17)  14(x + 17) = 0 (x + 17)(x  14) = 0 x = 17 or x = 14But the value x cannot be negative.Therefore, the breadth of the hall is 14 metre and the length is 17 metre.

Page No 495:

Question 61:

The perimeter of a rectangular plot is 62 m and its area is 228 sq metres. Find the dimensions of the plot.

Answer:

Let the length and breadth of the rectangular plot be x and y meter, respectively.Therefore, we have:Perimeter = 2(x + y) = 62               ...(i) andArea = xy = 228 y = 228xPutting the value of y in (i), we get⇒ 2(x + 228x) = 62 x + 228x = 31 x2 + 228x = 31 x2 + 228 = 31x x2  31x + 228 = 0 x2  (19 + 12)x + 228 = 0 x2  19x  12x + 228 = 0 x(x  19)  12(x  19) = 0 (x  19)(x  12) = 0 x = 19 or x = 12If x=19 m, y=22819 = 12 mTherefore, the length and breadth of the plot are 19 m and 12 m, respectively.

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Question 62:

A rectangular field is 16 m long and 10 m wide. There is a path of uniform width all around it, having an area of 120 m2. Find the width of the path.

Answer:

Let the width of the path be x m.  Length of the field including the path = 16 + x + x = 16 + 2xBreadth of the field including the path = 10 + x + x = 10 + 2xNow,(Area of the field including path)  (Area of the field excluding path) = Area of the path (16 + 2x)(10 + 2x)  (16 × 10) = 120 160 + 32x + 20x + 4x2  160 = 120 4x2 + 52x  120 = 0 x2 + 13x  30 = 0 x2 + (15  2)x + 30 = 0 x2 + 15x  2x + 30 = 0 x(x + 15)  2(x + 15) = 0 (x  2)(x + 15) = 0 x  2 = 0 or x + 15 = 0 x = 2 or x = 15 x = 2    ( Width cannot be negative)Thus, the width of the path is 2 m.

Page No 495:

Question 63:

The sum of the areas of two squares is 640 m2. If the difference in their perimeters be 64 m, find the sides of the two squares.

Answer:

Let the length of the side of the first and the second square be x and y, respectively.

According to the question:x2 + y2 = 640             ...(i)Also,4x  4y = 64 x  y = 16 x = 16 + y

Putting the value of x in (i), we get:

x2 + y2 = 640 (16 + y)2 + y2 = 640 256 + 32y + y2 + y2 = 640 2y2 + 32y  384 = 0 y2 + 16y  192 = 0 y2 + (24  8)y  192 = 0 y2 + 24y  8y  192 = 0 y(y + 24)  8(y + 24) = 0 (y + 24)(y  8) = 0 y = 24 or y = 8 y = 8    ( Side cannot be negative) x = 16 + y = 16 + 8 = 24 mThus, the sides of the squares are 8 m and 24 m.

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Question 64:

The length of a rectangle is thrice as long as the side of a square. The side of the square is 4 cm more than width of a the rectangle. Their areas being equal, find their dimensions.

Answer:

Let the breadth of rectangle be x cm.According to the question:Side of the square = (x + 4) cm Length of the rectangle = {3(x + 4)} cmIt is given that the areas of the rectangle and square are same. 3(x + 4) × x = (x + 4)2 3x2 + 12x = (x + 4)2 3x2 + 12x = x2 + 8x + 16 2x2 + 4x  16 = 0 x2 + 2x  8 = 0 x2 + (4  2)x  8 = 0 x2 + 4x  2x  8 = 0 x(x + 4)  2(x + 4) = 0 (x + 4)(x  2) = 0 x = 4 or x = 2 x = 2    ( The value of x cannot be negative)Thus, the breadth of the rectangle is 2 cm and length is {3(2 + 4) =18} cm. Also, the side of the square is 6 cm.

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Question 65:

A farmer prepares a rectangular vegetable garden of area 180 sq metres. With 39 metres of barbed wire, he can fence the three sides of the garden, leaving one of the longer sides unfenced. Find the dimensions of the garden.

Answer:

Let the length and breadth of the rectangular garden be x and y metre, respectively.Given:xy = 180 sq m       ...(i)   and2y + x = 39 x = 39  2yPutting the value of x in (i), we get:(39  2y)y = 180 39y  2y2 = 180 39y  2y2  180 = 0 2y2  39y + 180 = 0 2y2  (24 + 15)y + 180 = 0 2y2  24y  15y + 180 = 0 2y(y  12)  15(y  12) = 0 (y  12)(2y  15) = 0 y = 12 or y = 152 = 7.5If y = 12, x = 39  24 = 15If y = 7.5, x = 39  15 = 24Thus, the length and breadth of the garden are (15 m and 12 m) or (24 m and 7.5 m), respectively.



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Question 66:

The area of a right-triangle is  600 cm2. If the base of the triangle exceeds the altitude by 10 cm, find the dimensions of the triangle.

Answer:

Let the altitude of the triangle be x cm.
Therefore, the base of the triangle will be (+ 10) cm.
Area of triangle = 12 x (x + 10) = 600 x(x + 10) = 1200 x2 + 10x  1200 = 0 x2 + (40  30)x  1200 = 0 x2 + 40x  30x  1200 = 0 x(x + 40)  30(x + 40) = 0 (x + 40)(x  30) = 0 x = 40 or x = 30 x = 30   [ Altitude cannot be negative]Thus, the altitude and base of the triangle are 30 cm and (30 + 10 = 40) cm, respectively. Hypotenuse2 = Altitude2 + Base2 Hypotenuse2 = 302 + 402 Hypotenuse2 = 900 + 1600 = 2500 Hypotenuse2 = 502 Hypotenuse = 50Thus, the dimensions of the triangle are: ,Hypotenuse = 50 cmAltitude = 30 cmBase = 40 cm

Page No 496:

Question 67:

The area of a right-angled triangle is 96 sq metres. If the base is three times the altitude, find the base.

Answer:

Let the altitude of the triangle be x m.
Therefore, the base will be 3x m.

Area of a triangle = 12 × Base × Altitude

 12 × 3x × x = 96  ( Area = 96 sq m) x22 = 32 x2 = 64 x = ±8


The value of cannot be negative.
Therefore, the altitude and base of the triangle are 8 m and (3 × 8 = 24 m), respectively.

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Question 68:

The area of a right-angled triangle is 165 sq metres. Determine its base and altitude if the latter exceeds the former by 7 metres.

Answer:

Let the base be x m.
Therefore, the altitude will be x + 7 m.

Area of a triangle = 12 × Base × Altitude

 12 × x × (x + 7) = 165 x2 + 7x = 330 x2 + 7x  330 = 0 x2 + (22  15)x  330 = 0 x2 + 22x  15x  330 = 0 x(x + 22)  15(x + 22) = 0 (x + 22)(x  15) = 0 x = 22 or x = 15

The value of x cannot be negative.
Therefore, the base is 15 m and the altitude is {(15 + 7) = 22 m}.

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Question 69:

The hypotenuse of a right-angled triangle is 20 metres. If the difference between the length of the other sides be 4 metres, find the other sides.

Answer:

Let one side of the right-angled triangle be x m and the other side be x + 4 m.
On applying Pythagoras theorem, we have:

202 = (x + 4)2 + x2 400 = x2 + 8x + 16 + x2 2x2 + 8x  384 = 0 x2 + 4x  192 = 0 x2 + (16  12)x  192 = 0 x2 + 16x  12x  192 = 0 x(x + 16)  12(x + 16) = 0 (x + 16)(x  12) = 0 x = 16 or x = 12

The value of x cannot be negative.
Therefore, the base is 12 m and the other side is {(12 + 4) = 16 m}.

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Question 70:

The length of the hypotenuse of a right-angled triangle exceeds the length of the base by 2 cm and exceeds twice the length of the altitude by 1 cm. Find the length of each side of the triangle.

Answer:

Let the base and altitude of the right-angled triangle be x and y cm, respectively.
Therefore, the hypotenuse will be (x + 2) cm.
 (x + 2)2 = y2 + x2           ...(i)Again, the hypotenuse exceeds twice the length of the altitude by 1 cm. h = (2y + 1) x + 2 = 2y + 1 x = 2y  1Putting the value of x in (i), we get:(2y  1 +  2)2 = y2 + (2y  1)2 (2y + 1)2 = y2 + 4y2  4y + 1 4y2 + 4y + 1 = 5y2  4y + 1 y2 + 8y = 0 y2  8y = 0 y(y   8) = 0 y = 8 cm x = 16  1 = 15 cm h = 16 + 1 = 17 cm

Thus, the base, altitude and hypotenuse of the triangle are 15 cm, 8 cm and 17 cm, respectively.

Page No 496:

Question 71:

The hypotenuse of a right-angled triangle is 1 metre less than twice the shortest side. If the third is 1 metre more than the shortest side, find the sides of the triangle.

Answer:

Let the shortest side be x m.
Therefore, according to the question:
Hypotenuse = 2x - 1
Third side = x + 1 m
On applying Pythagoras theorem, we get:

(2x  1)2 = (x + 1)2 + x2 4x2  4x + 1 = x2 + 2x + 1 + x2 2x2  6x = 0 2x(x  3) = 0 x = 0 or x = 3

The length of the side cannot be 0; therefore, the shortest side is 3 m.
Therefore,
Hypotenuse = 2 × 3 - 1 = 5 m
Third side = (3 + 1) = 4 m



Page No 503:

Question 1:

Which of the following is a quadratic equation?
(a) x2-3x+2=0
(b) x+1x=x2
(c) x2+1x2=5
(d) 2x2  - 5x=(x-1)2

Answer:

(d) 2x2 -  5x = (x - 1)2

A quadratic equation is the equation with degree 2. 2x2  5x = (x  1)2 2x2  5x = x2  2x + 1 2x2  5x  x2 + 2x  1 = 0 x2  3x  1 = 0, which is a quadratic equation

Page No 503:

Question 2:

Which of the following is a quadratic equation?
(a) (x2+1)=(2-x)2+3
(b) x3-x2=(x-1)3
(c) 2x2+3=(5+x)(2x-3)
(d) None of these

Answer:

(b) x3 - x2 = (x - 1)3

∵ x3  x2 = (x  1)3⇒ x3  x2 = x3  3x2 + 3x  1⇒ 2x2  3x + 1 = 0, which is a quadratic equation



Page No 504:

Question 3:

Which of the following is not a quadratic equation?
(a) 3x-x2=x2+5
(b) (x+2)2=2(x2-5)
(c) (2x+3)2=2x2+6
(d) (x-1)2=3x2+x-2

Answer:

(c) (2x + 3)2 = 2x2 + 6

∵  (2x + 3)2 = 2x2 + 6⇒ 2x2 + 9 + 62x = 2x2 + 6⇒ 62x + 3 = 0, which is not a quadratic equation

Page No 504:

Question 4:

If x = 3 is a solution of the equation 3x2+(k-1)x+9=0, then k = ?
(a) 11
(b) −11
(c) 13
(d) −13

Answer:

(b) −11
It is given that x=3 is a solution of 3x2+(k1)x+9=0; therefore, we have:332 + k - 1 × 3 + 9 = 0 27 + 3(k  1) + 9 = 0 3(k  1) = 36 (k  1) = 12 k = 11

Page No 504:

Question 5:

If one root of the equation 2x2+ax+6=0 is 2, then a = ?
(a) 7
(b) −7
(c) 72
(d) -72

Answer:

(b) −7
It is given that one root of the equation 2x2 ax + 6 = 0 is 2. × 22 + a × 2 + 6 = 0 2a + 14 = 0 a = 7

Page No 504:

Question 6:

The sum of the roots of the equation x2-6x+2=0 is
(a) 2
(b) −2
(c) 6
(d) −6

Answer:

(c) 6
Sum of the roots of the equation x2  6x + 2 = 0 is α β ba = (6)1 = 6 , where α and β are the roots of the equation.

Page No 504:

Question 7:

If the product of the roots of the equation x2-3x+k=10 is −2, then the value of k is
(a) −2
(b) −8
(c)  8
(d) 12

Answer:

(c)  8
It is given that the product of the roots of the equation x2  3x + k = 10 is 2.The equation can be rewritten as:x2 - 3x + (k - 10) = 0Product of the roots of a quadratic equation = caca = 2 (k  10)1 = 2 k = 8

Page No 504:

Question 8:

The ratio of the sum and product of the roots of the equation 7x2-12x+18=0
(a) 7 : 12
(b) 7 : 18
(c) 2 : 3
(d) 3 : 2

Answer:

(c) 2 : 3

Given:7x2  12x + 18 = 0 α + β = 127 and αβ = 187, where α and β are the roots of the equation Ratio of the sum and product of the roots = 127 : 187= 12 : 18= 2  : 3

Page No 504:

Question 9:

If one root of the equation 3x2-10x+3=0 is 13, then the other root is
(a) -13
(b) 13
(c) −3
(d) 3

Answer:

(d) 3
Given: 3x2  10x + 3 = 0One root of the equation is 13.Let the other root be α.We know that:Product of the roots = ca 13 × α = 33 α = 3

Page No 504:

Question 10:

If one root of 5x2+13x+k=0 be the reciprocal of the other root, then the value of k is
(a) 0
(b) 1
(c) 2
(d) 5

Answer:

(d) 5
Let the roots of the equation (5x2 + 13x + k = 0) be α and 1α. Product of the roots = ca α × 1α = k5 1 = k5 k = 5

Page No 504:

Question 11:

If the sum of the roots of the equation kx2+2x+3x=0 is equal to their product, then the value of k is
(a) 13
(b) -13
(c) 23
(d) -23

Answer:

(d) -23
Given:kx2 + 2x + 3k = 0Sum of the roots = Product of the roots⇒ 2k = 3kk⇒ 3k = 2⇒ k = 23

Page No 504:

Question 12:

The root of a quadratic equation are 5 and −2. Then, the equation is
(a) x2-3x+10=0
(b) x2-3x-10=0
(c) x2+3x-10=0
(d) x2+3x+10=0

Answer:

(b) x2 - 3x - 10 = 0

It is given that the roots of the quadratic equation are 5 and 2.Then, the equation is:x2  (5  2)x + 5 × (2) = 0⇒ x2  3x  10 = 0

Page No 504:

Question 13:

If the sum of the roots of a quadratic equation is 6 and their product is 6, the equation is
(a) x2-6x+6=0
(b) x2+6x-6=0
(c) x2-6x-6=0
(d) x2+6x+6=0

Answer:

(a) x2 - 6x + 6 = 0

Given:Sum of roots = 6Product of roots = 6Thus, the equation is: x2  6x + 6 = 0



Page No 505:

Question 14:

Objective Questions (MCQ)

If α and β are the roots of the equation 3x2+8x+2=0 then 1α+1β = ?

(a) -38                            (b) 23                            (c) −4                            (d) 4

Answer:

It is given that α and β are the roots of the equation 3x2+8x+2=0.

 α+β=-83 and αβ=23
1α+1β=α+βαβ=-8323=-4

Hence, the correct answer is option C.

Page No 505:

Question 15:

The roots of the equation ax2+bx+c=0 will be reciprocal of each other if
(a) a = b
(b) bc
(c) c = a
(d) none of these

Answer:

(c) c  =  a
Let the roots of the equation (ax2 bx + c = 0) be α and 1α. Product of the roots = α × 1α = 1 ca = 1 c = a

Page No 505:

Question 16:

If the roots of the equation ax2+bx+c=0 are equal, then then c = ?
(a) -b2a
(b) b2a
(c) -b24a
(d) b24a

Answer:

(d) b24a

 It is given that the roots of the equation (ax2+bx+c=0) are equal. (b2  4ac) = 0 b2 = 4ac c = b24a

Page No 505:

Question 17:

If the equation  has equal roots, then k = ?
(a) 2 or 0
(b) −2 or 0
(c) 2 or −2
(d) 0 only

Answer:

(c) 2 or −2
It is given that the roots of the equation (9x2 6kx + 4 = 0) are equal. (b2  4ac) = 0 (6k)2  4 × 9 × 4 = 0 36k2 = 144 k2 = 4 k = ±2

Page No 505:

Question 18:

If the equation x2+2(k+2)x+9k=0 has equal rots, then k = ?
(a) 1 or 4
(b) −1 or 4
(c) 1 or −4
(d) −1 or −4

Answer:

(a) 1 or 4

It is given that the roots of the equation (x2 + 2(k + 2)x + 9k = 0) are equal. (b2  4ac) = 0 {2(k + 2)}2  4 × 1 × 9k = 0 4(k2 + 4k + 4)  36k = 0 4k2 + 16k + 16  36k = 0 4k2  20k + 16 = 0 k2  5k + 4 = 0 k2  4k  k + 4 = 0 k(k  4)  (k  4) = 0 (k  4)(k  1) = 0 k = 4 or k = 1

Page No 505:

Question 19:

If the equation 4x2-3kx+1=0 has equal roots, then k = ?
(a) ±23
(b) ±13
(c) ±34
(d) ±43

Answer:

(d) ±43

It is given that the roots of the equation (4x2  3kx + 1 = 0) are equal. (b2  4ac) = 0 (3k)2  4 × 4 × 1 = 0 9k2 = 16 k2 = 169 k = ±43

Page No 505:

Question 20:

The roots of ax2+bx+c=0, a0 are real and unequal, if (b2-4ac)
(a) > 0
(b) = 0
(c) < 0
(d) none of these

Answer:

(a) >  0
 The roots of the equation are real and unequal when (b2  4ac) > 0.

Page No 505:

Question 21:

Objective Questions (MCQ)

In the equation ax2+bx+c=0, it is given that D=b2-4ac>0. Then, the roots of the equation are

(a) real and equal                                                        (b) real and unequal
(c) imaginary                                                              (d) none of these

Answer:

We know that when discriminant, D>0, the roots of the given quadratic equation are real and unequal.

Hence, the correct answer is option B.

Page No 505:

Question 22:

The roots of the equation 2x2-6x+7=0 are
(a) real, unequal and rational
(b) real, unequal and irrational
(c) real and equal
(d) imaginary

Answer:

(d) imaginary

 D = (b2  4ac)= (6)2  4 × 2 × 7= 36  56= 20 < 0Thus, the roots of the equation are imaginary.

Page No 505:

Question 23:

The roots of the equation 2x2-6x+3=0 are
(a) real, unequal and rational
(b) real, unequal and irrational
(c) real and equal
(d) imaginary

Answer:

(b) real, unequal and irrational

 D = (b2  4ac)= (6)2  4 × 2 × 3= 36  24= 12 12 is greater than 0 and it is not a perfect square; therefore, the roots of the equation are real, unequal and irrational.

Page No 505:

Question 24:

If the roots of 5x2-kx+1=0 are real and distinct, then
(a) -25<k<25
(b) k>25 only
(c) k<-25 only
(d) either k>25 or k<-25

Answer:

(d) either k > 25 or k < -25

It is given that the roots of the equation (5x2  kx + 1 = 0) are real and distinct.(b2  4ac) > 0 (k)2  4 × 5 × 1 > 0 k2  20 > 0 k2 > 20 k > 20 or k < 20 k > 25 or k < 25

Page No 505:

Question 25:

If the equation x2+5kx+16=0 has no real roots, then
(a) k>85
(b) k<-85
(c) -85<k>85
(d) none of these

Answer:

(c) -85 < k > 85

It is given that the equation (x2 + 5kx + 16 = 0) has no real roots. (b2  4ac) < 0 (5k)2  4 × 1 × 16 < 0 25k2  64 < 0 k2 < 6425 85 < k < 85



Page No 506:

Question 26:

If the equation x2-kx+1=0 has no real roots, then
(a) k < −2
(b) k > 2
(c) −2 < k < 2
(d) none of these

Answer:

(c) −2  <   <  2

It is given that the equation x2  kx + 1 = 0 has no real roots. (b2  4ac) < 0 (k)2  4 × 1 × 1 < 0 k2 < 4 2 < k < 2

Page No 506:

Question 27:

For what values of k, the equation kx2-6x-2=0 has real roots?
(a) k-92
(b) k-92
(c) k-2
(d) None of these

Answer:

(b) k  -92

It is given that the roots of the equation (kx2  6x  2 = 0) are real. D  0 (b2  4ac)  0 (6)2  4 × k × (2)  0 36 + 8k  0 k  368 k  92

Page No 506:

Question 28:

The sum of a number and its reciprocal is 2120. The number is
(a) 54 or 45
(b) 43 or 34
(c) 56 or 65
(d) 16 or 6

Answer:

(a) 54 or 45

Let the required number be x.According to the question: x + 1x = 4120 x2 + 1x = 4120 20x2  41x + 20 = 0 20x2  25x  16x + 20 = 0 5x(4x  5)  4(4x  5) = 0 (4x  5)(5x  4) = 0 x = 54 or x = 45

Page No 506:

Question 29:

The perimeter of a rectangle is 82 m and its area is 400 m2. The breadth of the rectangle is
(a) 25 m
(b) 20 m
(c) 16 m
(d) 9 m

Answer:

(c) 16 m

Let the length and breadth of the rectangle be l and b.Perimeter of the rectangle = 82 m 2 × (l + b) = 82 l + b = 41 l = (41  b)           ...(i)Area of the rectangle = 400 m2 l × b = 400 m2 (41  b)b = 400      (using (i)) 41b  b2 = 400 b2  41b + 400 = 0 b2  25b  16b + 400 = 0 b(b  25)  16(b  25) = 0 (b  25)(b  16) = 0 b = 25 or b = 16If b = 25, we have:l = 41  25 = 16 Since, l cannot be less than b, b = 16 m

Page No 506:

Question 30:

Objective Questions (MCQ)

The length of a rectangular field exceeds its breadth by 8 m and the area of the field is 240 m2. The breadth of the field is

(a) 20 m                                (b) 30 m                                (c) 12 m                                (d) 16 m                                           [CBSE 2014]

Answer:

Let the breadth of the rectangular field be x m.

∴ Length of the rectangular field = (x + 8) m

Area of the rectangular field = 240 m2                    (Given)

x+8×x=240                       (Area = Length × Breadth)
x2+8x-240=0x2+20x-12x-240=0xx+20-12x+20=0x+20x-12=0
x+20=0 or x-12=0x=-20 or x=12
x = 12              (Breadth cannot be negative)

Thus, the breadth of the field is 12 m.

Hence, the correct answer is option C.

Page No 506:

Question 31:

Objective Questions (MCQ)

The roots of the quadratic equation 2x2-x-6=0 are

(a) -2, 32                                    (b) 2, -32                                    (c) -2, -32                                    (d) 2, 32                     [CBSE 2012]                           

Answer:

The given quadratic equation is 2x2-x-6=0.

2x2-x-6=02x2-4x+3x-6=02xx-2+3x-2=0x-22x+3=0
x-2=0 or 2x+3=0x=2 or x=-32
Thus, the roots of the given equation are 2 and -32.

Hence, the correct answer is option B.

Page No 506:

Question 32:

The sum of two natural numbers is 8 and their product is 15. Find the numbers.                                          [CBSE 2012]

Answer:

Let the required natural numbers be x and (8 − x).

It is given that the product of the two numbers is 15.
x8-x=158x-x2=15x2-8x+15=0x2-5x-3x+15=0
xx-5-3x-5=0x-5x-3=0x-5=0 or x-3=0x=5 or x=3

Hence, the required numbers are 3 and 5.

Page No 506:

Question 33:

Very-Short-Answer Questions

Show that x = −3 is a solution of x2+6x+9=0.                         [CBSE 2008]

Answer:

The given equation is x2+6x+9=0.

Putting x = −3 in the given equation, we get

LHS = -32+6×-3+9=9-18+9=0 = RHS

∴ x = −3 is a solution of the given equation.

Page No 506:

Question 34:

Very-Short-Answer Questions

Show that = −2 is a solution of 3x2+13x+14=0.                          [CBSE 2008]

Answer:

The given equation is 3x2+13x+14=0.

Putting x = −2 in the given equation, we get

LHS = 3×-22+13×-2+14=12-26+14=0 = RHS

∴ x = −2 is a solution of the given equation.

Page No 506:

Question 35:

Very-Short-Answer Questions

If x=-12 is a solution of the quadratic equation 3x2+2kx-3=0, find the value of k.                     [CBSE 2015]

Answer:

It is given that x=-12 is a solution of the quadratic equation 3x2+2kx-3=0.

3×-122+2k×-12-3=034-k-3=0k=3-124=-94

Hence, the value of k is -94.

Page No 506:

Question 36:

Very-Short-Answer Questions

Find the roots of the quadratic equation 2x2-x-6=0.                                 [CBSE 2012]

Answer:

The given quadratic equation is 2x2-x-6=0.
2x2-x-6=02x2-4x+3x-6=02xx-2+3x-2=0x-22x+3=0
x-2=0 or 2x+3=0x=2 or x=-32
Hence, the roots of the given equation are 2 and -32.

Page No 506:

Question 37:

Very-Short-Answer Questions

Find the solution of the quadratic equation 33x2+10x+3=0.                    [CBSE 2009]

Answer:

The given quadratic equation is 33x2+10x+3=0.
33x2+10x+3=033x2+9x+x+3=033xx+3+1x+3=0x+333x+1=0
x+3=0 or 33x+1=0x=-3 or x=-133=-39
Hence, -3 and -39 are the solutions of the given equation.

Page No 506:

Question 38:

Very-Short-Answer Questions

If the roots of the quadratic equation 2x2+8x+k=0 are equal then find the value of k.                     [CBSE 2014]

Answer:

It is given that the roots of the quadratic equation 2x2+8x+k=0 are equal.
D=082-4×2×k=064-8k=0k=8
Hence, the value of k is 8.

Page No 506:

Question 39:

Very-Short-Answer Questions

If the quadratic equation px2-25px+15=0 has two equal roots then find the value of p.                   [CBSE 2015]

Answer:

It is given that the quadratic equation px2-25px+15=0 has two equal roots.

D=0-25p2-4×p×15=020p2-60p=020pp-3=0
p=0 or p-3=0p=0 or p=3

For p = 0, we get 15 = 0, which is not true.

p ≠ 0

Hence, the value of p is 3.

Page No 506:

Question 40:

Very-Short-Answer Questions

If 1 is a root of the equation ay2+ay+3=0 and y2+y+b=0 then find the value of ab.                 [CBSE 2012]

Answer:

It is given that y = 1 is a root of the equation ay2+ay+3=0.
a×12+a×1+3=0a+a+3=02a+3=0a=-32

Also, y = 1 is a root of the equation y2+y+b=0.
12+1+b=01+1+b=0b+2=0b=-2

ab=-32×-2=3

Hence, the value of ab is 3.



Page No 507:

Question 41:

Very-Short-Answer Questions

If one zero of the polynomial x2-4x+1 is 2+3, write the other zero.                    [CBSE 2010]

Answer:

Let the other zero of the given polynomial be α.

Now,
Sum of the zeroes of the given polynomial = --41= 4
 α+2+3=4α=4-2-3=2-3

Hence, the other zero of the given polynomial is 2-3.

Page No 507:

Question 42:

Very-Short-Answer Questions

If one root of the quadratic equation 3x2-10x+k=0 is reciprocal of the other, find the value of k.                  [CBSE 2014]

Answer:

Let α and β be the roots of the equation 3x2-10x+k=0.
 α=1β          (Given)
αβ=1k3=1            Product of the roots=cak=3

Hence, the value of k is 3.

Page No 507:

Question 43:

Very-Short-Answer Questions

If the roots of the quadratic equation pxx-2+6=0 are equal, find the value of p.                         [CBSE 2013]

Answer:

It is given that the roots of the quadratic equation px2-2px+6=0 are equal.
D=0-2p2-4×p×6=04p2-24p=04pp-6=0
p=0 or p-6=0p=0 or p=6

For p = 0, we get 6 = 0, which is not true.

p ≠ 0

Hence, the value of p is 6.

Page No 507:

Question 44:

Very-Short-Answer Questions

Find the values of k so that the quadratic equation x2-4kx+k=0 has equal roots.                         [CBSE 2012]

Answer:

It is given that the quadratic equation x2-4kx+k=0 has equal roots.
 D=0-4k2-4×1×k=016k2-4k=04k4k-1=0
k=0 or 4k-1=0k=0 or k=14

Hence, 0 and 14 are the required values of k.

Page No 507:

Question 45:

Very-Short-Answer Questions

Find the values of k for which the quadratic equation 9x2-3kx+k=0 has equal roots.                         [CBSE 2014]

Answer:

It is given that the quadratic equation 9x2-3kx+k=0 has equal roots.