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Page No 36:

Question 1:

Choose the correct answer from the given four options.
Which of the following is a quadratic equation?
(A) x2 + 2x + 1 = (4 – x)2 + 3
(B) -2x2=5-x2x-25
(C) k+1x2+32x=7, where k=-1
(D) x3 x2 = (x – 1)3

Answer:

An equation is of the form ax2+bx+c=0a0 is called a quadratic equation. So, simplify each part of the question and check whether it is in the form of ax2+bx+c=0 or not.

(a) Given that, 
x2+2x+1=(4-x)2+3
x2+2x+1=16+x2-8x+3
10x-18=0, which is not a quadratic equation.

(b) Given that,
-2x2=(5-x)2x-25
-2x2=10x-2x2-2+2x5
50x+2x-10=0
52x-10=0, which is not a quadratic equation.

(c) Given that,
x2k+1+32x=7, where k = −1
x2-1+1+32x=7
3x-14=0, which is not a quadratic equation.

(d) Given that,
x3-x2=x-13
x3-x2=x3-3x2+3x-1-x2+3x2-3x+1=0
2x2-3x+1=0, which is a quadratic equation.

Hence, the correct answer is option D.

 

Page No 36:

Question 2:

Choose the correct answer from the given four options.
Which of the following is not a quadratic equation?
(A) 2(x – 1)2 = 4x2 – 2x + 1
(B) 2xx2 = x2 + 5
(C) 2x+32+x2=3x2-5x
(D) (x2 + 2x)2 = x4 + 3 + 4x3

Answer:

An equation is of the form ax2+bx+c=0a0 is called a quadratic equation. So, simplify each part of the question and check whether it is in the form of ax2+bx+c=0 or not.

(a) Given that,
2x-12=4x2-2x+1
2x2+1-2x=4x2-2x+12x2+2-4x=4x2-2x+1
2x2+2x-1=0, which is a quadratic equation.

(b) Given that,
2x-x2=x2+5
2x2-2x+5=0, which is a quadratic equation.

(c) Given that,
2x+32+x2=3x2-5x
2x2+3+26x+x2=3x2-5x
5x+3+26x=0, which is not a quadratic equation.

(d) Given that,
x2+2x2=x4+3+4x3
x4+4x2+4x3=x4+3+4x3
4x2-3=0, which is a quadratic equation.

Hence, the correct answer is option C.

Page No 36:

Question 3:

Choose the correct answer from the given four options.
Which of the following equations has 2 as a root?
(A) x2 – 4x + 5 = 0
(B) x2 + 3x – 12 = 0
(C) 2x2 – 7x + 6 = 0
(D) 3x2 – 6x – 2 = 0

Answer:

If α is one of the root of quadratic equation ax2+bx+c=0, then x = α  satisfies the equation aα2+bα+c=0.

(a) Given, x2-4x+5=0
Substituting x = 2 in x2-4x+5, we get 
22-42+5
4-8+5=10, So we conclude, x = 2 is not a root of x2-4x+5=0.

(b) Given, x2+3x-12=0
Substituting x = 2 in x2+3x-12, we get
22+32-12
4+6-12=-20, So we conclude, x = 2 is not a root of x2+3x-12=0.

(c) Given, 2x2-7x+6=0
Substituting x = 2 in 2x2-7x+6, we get
2(2)2-7(2)+6
8-14+6=14-14=0, So we conclude, x = 2 is a root of 2x2-7x+6=0.

(d) Given, 3x2-6x-2=0
Substituting x = 2 in 3x2-6x-2, we get
3(2)2-6(2)-2
12-12-2=-20, So we conclude, x = 2 is not a root of 3x2-6x-2=0.

Hence, the correct answer is option C.



Page No 37:

Question 4:

Choose the correct answer from the given four options.
If 12is a root of the equation x2+kx-54 = 0, then the value of k is
(A) 2

(B) – 2

(C) 14

(D)12

Answer:

Since, 12 is a root of given equation, then put x=12 in given equation and get the value of k.

Given, 
12 is a root of the quadratic equation x2+kx-54=0.
122+k12-54=014+k2-54=01+2k-54=0
2k-4=02k=4k=2

Thus, we conclude the value of k = 2.
Hence, the correct answer is option A.

Page No 37:

Question 5:

Choose the correct answer from the given four options.
Which of the following equations has the sum of its roots as 3?
(A) 2x2 – 3x + 6 = 0
(B) –x2 + 3x – 3 = 0
(C) 2x2-32x+1=0
(D) 3x2 – 3x + 3 = 0

Answer:

If α and β are the roots of the quadratic equation ax2+bx+c=0a0, then sum of roots = α + β = -1Coefficient of xCoefficient of x2=(-1)ba.

(a) Given, 2x2-3x+6=0.
Comparing with ax2+bx+c=0, we get
a = 2, b = −3 and c = 6.
∴ Sum of the roots = -ba=--32=32.
So, we conclude sum of the roots of given quadratic equation is not 3.

(b) Given, -x2+3x-3=0.
Comparing with ax2+bx+c=0, we get
a = −1, b = 3 and c = −3.
∴Sum of the roots = -ba=-3-1=3.
So, we conclude sum of the roots of given quadratic equation is 3.

(c) Given, 2x2-32x+1=0.
Comparing with ax2+bx+c=0, we get
a = 2, b = −3 and c = 2.
∴Sum of the roots = -ba=--32=32.
So, we conclude sum of the roots of given quadratic equation is not 3.

(d) Given, 3x2-3x+3=0.
Comparing with ax2+bx+c=0, we get
a = 1, b = −1 and c = 1.
∴Sum of the roots = -ba=--11=1.
So, we conclude sum of the roots of given quadratic equation is not 3.

Hence, the correct answer is option B.

 

Page No 37:

Question 6:

Choose the correct answer from the given four options.
Values of k for which the quadratic equation 2x2kx + k = 0 has equal roots is
(A) 0 only
(B) 4
(C) 8 only
(D) 0, 8

Answer:

Given, 2x2-kx+k=0
Now, comparing with ax2+bx+c=0, we get
a = 2, b = −k and c = k.

For equal roots, the discriminant must be zero.
i.e. we can say, D=b2-4ac=0.
-k2-42k=0k2-8k=0kk-8=0

k = 0, 8

Thus, the required values of k are 0 and 8.
Hence, the correct answer is option D.

Page No 37:

Question 7:

Choose the correct answer from the given four options.
Which constant must be added and subtracted to solve the quadratic equation 9x2+34x-2=0 by the method of completing the square?
(A) 18

(B) 164

(C) 14

(D) 964

Answer:

Given, equation 9x2+34x-2=0.
3x2+143x-2=0
Substituting 3x = y in above equation, we get
y2+14y-2=0
y2+14y+182-182-2=0y+182=164+2y+182=1+64264

Thus, 164 must be added and subtracted to solve the given equation.
Hence, the correct answer is option B.

Page No 37:

Question 8:

Choose the correct answer from the given four options.
The quadratic equation 2x2-5x+1=0 has
(A) two distinct real roots
(B) two equal real roots
(C) no real roots
(D) more than 2 real roots

Answer:

Given, equation 2x2-5x+1=0.
Now, comparing with ax2+bx+c=0, we get
a = 2, -5 and c = 1.
D=b2-4ac=-52-4×2×1=5-8, where D is discriminant.
D=-3, which is less than 0.

Thus, we conclude given quadratic equation has no real roots since discriminant is negative.
Hence, the correct answer is option C.

Page No 37:

Question 9:

Choose the correct answer from the given four options.
Which of the following equations has two distinct real roots?
(A) 2x2-32x+94=0
(B) x2 + x – 5 = 0
(C) x2+3x+22=0
(D) 5x2 – 3x + 1 = 0

Answer:

If a quadratic equation is in the form of ax2+bx+c=0a0 then
(i) If D=b2-4ac>0, then its roots are distinct and real.
(ii) If D=b2-4ac=0, then its roots are real and equal.
(iii) If D=b2-4ac<0, then its roots are not real or imaginary roots.

(a) Given, 2x2-32x+94=0.
Now, comparing with ax2+bx+c=0, we get
a = 2, -32 and c = 94.
D=b2-4ac=-322-4×2×94=18-18=0, where D is discriminant.
D=0.

Thus, equation has real and equal roots.

(b) Given x2+x-5=0.
Now, comparing with ax2+bx+c=0, we get
a = 1, = 1 and c = −5.
D=b2-4ac=12-4×1×-5=1+20=21, where D is discriminant.
D=21, which is greater than zero.

Thus, equation has real and distinct roots.

(c) Given x2+3x+22=0.
Now, comparing with ax2+bx+c=0, we get
a = 1, = 3 and c = 22
D=b2-4ac=32-4×1×22=9-82<0, where D is discriminant.
D<0

Thus, equation has no real roots.

(d) Given 5x2-3x+1=0.
Now, comparing with ax2+bx+c=0, we get
a = 5, = −3 and c = 1
D=b2-4ac=-32-4×5×1=9-20<0, where D is discriminant.
D<0

Thus, equation has no real roots.

Hence, the correct answer is option B.
 

Page No 37:

Question 10:

Choose the correct answer from the given four options.
Which of the following equations has no real roots?
(A) x2-4x+32=0
(B) x2+4x-32=0
(C) x2-4x-32=0
(D) 3x2+43x+4=0

Answer:

If a quadratic equation is in the form of ax2+bx+c=0a0 then
(i) If D=b2-4ac>0, then its roots are distinct and real.
(ii) If D=b2-4ac=0, then its roots are real and equal.
(iii) If D=b2-4ac<0, then its roots are not real or imaginary roots.

(a) Given x2-4x+32=0.
Now, comparing with ax2+bx+c=0, we get
a = 1, = −4 and c = 32.
D=b2-4ac=-42-4×1×32=16-122<0, where D is discriminant.
D<0.

Thus, equation has no real roots.

(b) Given x2+4x-32=0.
Now, comparing with ax2+bx+c=0, we get
a = 1, = 4 and c = -32.
D=b2-4ac=42-4×1×(-32)=16+122>0, where D is discriminant.
D>0.

Thus, equation has real and distinct roots.

(c) Given x2-4x-32=0.
Now, comparing with ax2+bx+c=0, we get
a = 1, = −4 and c = -32.
D=b2-4ac=-42-4×1×(-32)=16+122>0, where D is discriminant.
D>0.

Thus, equation has real and distinct roots.

(d) Given 3x2+43x+4=0.
Now, comparing with ax2+bx+c=0, we get
a = 3, 43 and c = 4.
D=b2-4ac=432-4×3×4=48-48=0, where D is discriminant.
D=0.

Thus, equation has real and equal roots.

Hence, the correct answer is option A.



Page No 38:

Question 11:

Choose the correct answer from the given four options.
(x2 + 1)2x2 = 0 has
(A) four real roots
(B) two real roots
(C) no real roots
(D) one real root.

Answer:

Given equation, x2+12-x2=0.
x4+1+2x2-x2=0x4+x2+1=0

Let x2=y
∴ x22+x2+1=0
y2+y+1=0

Now, comparing with ay2+by+c=0, we get.
a = 1, b = 1 and c = 1
Thus, we can say D=b2-4ac, where D is discriminant.
=12-411=1-4=-3

Since, D<0.
Thus, we conclude given equation has no real roots.
Hence, the correct answer is option C.

Page No 38:

Question 1:

State whether the following quadratic equations have two distinct real roots. Justify your answer.
(i) x2 – 3x + 4 = 0
(ii) 2x2 + x – 1 = 0
(iii) 2x2-6x+92=0
(iv) 3x2 – 4x + 1 = 0
(v) (x + 4)2 – 8x = 0
(vi) x-22-2x+1=0
(vii) 2x2-32x+12=0
(viii) x(1 – x) – 2 = 0
(ix) (x – 1) (x + 2) + 2 = 0
(x) (x + 1) (x – 2) + x = 0

Answer:

If a quadratic equation is in the form of ax2+bx+c=0a0 then
(i) If D=b2-4ac>0, then its roots are distinct and real.
(ii) If D=b2-4ac=0, then its roots are real and equal.
(iii) If D=b2-4ac<0, then its roots are not real or imaginary roots.

(i) Given x2-3x+4=0.
Now, comparing with ax2+bx+c=0, we get
a = 1, b = −3 and c = 4
D=b2-4ac=-32-414=9-16=-7, where D is discriminant.
D<0.
Thus, equation has no real roots.

(ii) Given 2x2+x-1=0.
Now, comparing with ax2+bx+c=0, we get
= 2, b = 1 and c = −1
 D=b2-4ac=12-42-1=1+8=9, where D is discriminant.
 D>0.
​Thus, equation has two distinct and real roots.

(iii) Given 2x2-6x+92=0.
Now, comparing with ax2+bx+c=0, we get
= 2, b = −6 and c = 92
D=b2-4ac=-62-4292=36-36=0, where D is discriminant.
D=0.
​Thus, equation has equal and real roots.

(iv) Given 3x2-4x+1=0.
Now, comparing with ax2+bx+c=0, we get
= 3, b = −4 and c = 1
D=b2-4ac=-42-431=16-12=4, where D is discriminant.
D>0.
Thus, equation has two distinct real roots.

(v) Given x+42-8x=0.
x2+16+8x-8x=0x2+16=0x2+0x+16=0
Now, comparing with ax2+bx+c=0, we get
= 1, b = 0 and c = 16
D=b2-4ac=02-4116=-64, where D is discriminant.
D<0.
Thus, equation has no real roots.

(vi) Given  x-22-2(x+1)=0.
x2+22-2x2-2x-2=0x2+2-22x-2x-2=0x2-x22+2=0x2-x22+2+0=0
Now, comparing with ax2+bx+c=0, we get
= 1, b = -22+2 and c = 0
D=b2-4ac=-22+22-410, where D is discriminant.
D=22+22
D>0.
Thus, equation has two distinct real roots.

(vii) Given equation, 2x2-32x+12=0
Now, comparing with ax2+bx+c=0, we get
= 2b = -32 and c = 12
D=b2-4ac=-322-4212=92-4=12, where D is discriminant.
D>0.
Thus, equation has two distinct real roots.

(viii) Given equation x1-x-2=0.
x-x2-2=0x2-x+2=0
Now, comparing with ax2+bx+c=0, we get
= 1, b = −1 and c = 2
D=b2-4ac=-12-412=1-8=-7, where D is discriminant.
D<0.
Thus, equation has no real roots.

(ix) Given equation x-1x+2+2=0.
x2+x-2+2=0x2+x+0=0
Now, comparing with ax2+bx+c=0, we get
= 1, b = 1 and c = 0
D=b2-4ac=12-410=1, where D is discriminant.
D>0.
Thus, equation has two distinct and real roots.

(x) Given equation x+1x-2+x=0.
x2+x-2x-2+x=0x2-2=0x2+0x-2=0
Now, comparing with ax2+bx+c=0, we get
= 1, b = 0 and c = −2
D=b2-4ac=02-41-2=0+8=8, where D is discriminant.
D>0.
Thus, equation has two distinct real roots.








 

Page No 38:

Question 2:

Write whether the following statements are true or false. Justify your answers.
(i) Every quadratic equation has exactly one root.
(ii) Every quadratic equation has at least one real root.
(iii) Every quadratic equation has at least two roots.
(iv) Every quadratic equations has at most two roots.
(v) If the coefficient of x2 and the constant term of a quadratic equation have opposite signs, then the quadratic equation has real roots.
(vi) If the coefficient of x2 and the constant term have the same sign and if the coefficient of x term is zero, then the quadratic equation has no real roots.

Answer:

(i) False, since a quadratic equation has exactly two roots.
(ii) False, since consider equation x2+4=0, has no real roots.
(iii) False, since a quadratic has two and only two roots.
(iv) False, every quadratic equation has exactlytwo roots.
(v) True
Consider, D=b2-4ac
ac<0 and b2>0
Thus, discriminant will always be positive.
(vi) True
Consider, D=b2-4ac
Now, if b = 0, then b2-4ac=-4ac and ac > 0.
Thus, discriminant will always be negative.



Page No 39:

Question 3:

A quadratic equation with integral coefficient has integral roots. Justify your answer.

Answer:

No
Consider the quadratic equation 5x2-3x-8=0 with integral coefficient.
Now lets solve the quadratic equation.
5x2-3x-8=0.
5x2-8x+5x-8=0x5x-8+1(5x-8)=0(5x-8)(x+1)=0x=85,-1
Thus, the roots of quadratic equation are not integers.
Hence, we conclude roots can be integral, but not always integral.

Page No 39:

Question 4:

Does there exist a quadratic equation whose coefficients are rational but both of its roots are irrational? Justify your answer.

Answer:

Yes
Consider the quadratic equation 2x2-3x-15=0, where coefficients are rational.
Now lets solve the quadratic equation using quadratic formula -b±b2-4ac2a, where a, b and c coefficients of the quadratic equation ax2+bx+c=0.
Comparing the equation, we get
a = 2, b = −3 and c = −15 
D=b2-4ac=(-3)2-4(2)(-15)=9+120=129, where D is discriminant.
Thus, roots are given 
x=-(-3)±1292×2
x=3±1294
Thus, we conclude roots are irrational since 129 is irrational.
 

Page No 39:

Question 5:

Does there exist a quadratic equation whose coefficients are all distinct irrationals but both the roots are rationals? Why?

Answer:

Yes
Consider the quadratic equation 3x2-73x+123=0, where coefficients are distinct irrational.
Now lets solve the quadratic equation using quadratic formula -b±b2-4ac2a, where ab and c  are coefficients of the quadratic equation ax2+bx+c=0.
Comparing the equation, we get
a = 3b = -73 and c 123
D=b2-4ac=(-73)2-4(3)(123)
D=49×3-4×12×3=349-48
D=3, where D is discriminant.

Thus, roots are given
x=73±323
x=73+323 or 73-323
x=37+123or37-123
Solving further, we get
x=4, 3

Thus, we conclude roots are rational.

Page No 39:

Question 6:

Is 0.2 a root of the equation x2 – 0.4 = 0? Justify.

Answer:

Given equation, x2-0.4=0
Now, substitute the value 0.2 in the given equation, we get
0.22-0.4=0.04-0.40

Thus, we conclude 0.2 is not a root of the given equation.

Page No 39:

Question 7:

If b = 0, c < 0, is it true that the roots of x2 + bx + c = 0 are numerically equal and opposite in sign? Justify.

Answer:

Given, b = 0 and c < 0, for the quadratic equation x2+bx+c=0.
Now, substitute b = 0 in the given equation, we get
x2+0+c=0x2+c=0x2=-c

∴ x=±-c

Thus, we conclude the roots of the given quadratic equation are numerically equal and opposite in sign.
 



Page No 40:

Question 1:

Find the roots of the quadratic equations by using the quadratic formula in each of the following:
(i) 2x2 – 3x – 5 = 0
(ii) 5x2 + 13x + 8 = 0
(iii) –3x2 + 5x + 12 = 0
(iv) –x2 + 7x – 10 = 0
(v) x2+22x-6=0
(vi) x2-35x+10=0
(vii) 12x2-11x+1=0

Answer:

(i) Given equation, 2x2-3x-5=0.
Now, comparing the equation with ax2+bx+c=0, we get
a = 2, b = −3 and c = −5

This implies that using quadratic formula x=-b±b2-4ac2a , we get
=--3±-32-4(2)(-5)2(2)
=3±9+404=3±494
=3±74=104,-44=52,-1

Thus, the roots of the given quadratic equation are 52 and -1.

(ii) Given equation, 5x2+13x+8=0.
Now, comparing the equation with ax2+bx+c=0, we get
a = 5, b = 13 and = 8

This implies that using quadratic formula x=-b±b2-4ac2a , we get
=-13±132-4(5)(8)2(5)
=-13±169-16010=-13±910
=-13±310=-1010,-1610=-1,-85

Thus, the roots of the given quadratic equation are -85 and -1.

(iii) Given equation, -3x2+5x+12=0.
Now, comparing the equation with ax2+bx+c=0, we get
a = −3, b = 5 and = 12

This implies that using quadratic formula x=-b±b2-4ac2a , we get
=-5±52-4(-3)(12)2(-3)
=-5±25+144-6=-5±169-6
=-5±13-6=8-6,-18-6=-43,3

Thus, the roots of the given quadratic equation are -43 and 3.


(iv) Given equation, -x2+7x-10=0.
Now, comparing the equation with ax2+bx+c=0, we get
a = −1, b = 7 and = −10

This implies that using quadratic formula x=-b±b2-4ac2a , we get
=-7±72-4(-1)(-10)2(-1)
=-7±49-40-2=-7±9-2
=-7±9-2=-4-2,-10-2=2,5

Thus, the roots of the given quadratic equation are 2 and 5.


(v) Given equation, x2+22x-6=0.
Now, comparing the equation with ax2+bx+c=0, we get
a = 1, b = 22 and = −6

This implies that using quadratic formula x=-b±b2-4ac2a , we get
=-22±222-4(1)(-6)2(1)
=-22±8+242=-22±322
=2,-32

Thus, the roots of the given quadratic equation are 2 and -32.


(vi) Given equation, x2-35x+10=0.
Now, comparing the equation with ax2+bx+c=0, we get
a = 1, b = -35 and = 10

This implies that using quadratic formula x=-b±b2-4ac2a , we get
=--35±-352-4(1)(10)2(1)
=35±45-402=35±52
=25,5

Thus, the roots of the given quadratic equation are 25 and 5.


(vii) Given equation, 12x2-11x+1=0.
Now, comparing the equation with ax2+bx+c=0, we get
a = 12b = -11 and = 1

This implies that using quadratic formula x=-b±b2-4ac2a , we get
=--11±-112-412(1)212
=11±11-21=11±9
=3+11, 11-3

Thus, the roots of the given quadratic equation are 3+11 and 11-3.

 

 

Page No 40:

Question 2:

Find the roots of the following quadratic equations by the factorisation method:
(i) 2x2+53x-2=0

(ii) 25x2-x-35=0

(iii) 32x2-5x-2=0

(iv) 3x2+55x-10=0

(v) 21x2-2x+121=0

Answer:

(i) Given 2x2+53x-2=0.
Multiplying both sides by 3, we get
6x2+5x-6=06x2+9x-4x-6=06x2+9x-4x-6=03x(2x+3)-2(2x+3)=0(2x+3)(3x-2)=0

Now, 2x+3=0
x=-32
and 3x-2=0
x=23
Hence, the roots of the equation are -32 and 23.

(ii) Given 25x2-x-35=0.
Multiplying both sides by 5, we get
2x2-5x-3=0
2x2-(6x-x)-3=02x2-6x+x-3=02x(x-3)+1(x-3)=0(x-3)(2x+1)=0

Now, x-3=0
x=3
and 2x+1=0
x=-12
Hence, the roots of the given equation are -12and 3.

(iii) Given 32x2-5x-2=0.
32x2-6x-x-2=032x2-6x+x-2=032x2-(32)(2x)+x-2=0
32x(x-2)+1(x-2)=0(x-2)(32x+1)=0

Now, x-2=0
x=2
and 32x+1=0
x=-132=-26

Hence, the roots of the equation are -26 and 2.

(iv) Given  3x2+55x-10=0.
3x2+65x-5x-10=0
3x2+65x-5x-2(5)(5)=0
3x(x+25)-5(x+25)=0(x+25)(3x-5)=0

Now, x+25=0
x=-25
and 3x-5=0
x=53

Hence, the roots of the equation are -25 and 53.

(v) Given 21x2-2x+121=0.
Multiplying both sides by 21, we get
441x2-42x+1=0441x2-(21x+21x)+1=0441x2-21x-21x+1=021x(21x-1)-1(21x-1)=0(21x-1)(21x-1)=0

Now, 21x-1=0
x=121
and 21x-1=0
x=121

Hence, the roots of the quadratic equation are 121 and 121.
 



Page No 42:

Question 1:

Find whether the following equations have real roots. If real roots exist, find them.
(i) 8x2 + 2x – 3 = 0

(ii) –2x2 + 3x + 2 = 0

(iii) 5x2 – 2x – 10 = 0

(iv) 12x-3+1x-5=1, x32, 5

(v) x2+55x-70=0

Answer:

(a) Firstly we will check the quadratic equation has real roots or not, for this we will verify the discriminant 
If discriminant, D=b2-4ac0 then roots are real.
(b) If roots are real, then further we can factorise the equation or use the quadratic formula to obtain the roots of the equation.

(i) Given 8x2+2x-3=0.
Now, comparing with ax2+bx+c=0, we get
a = 8, b = 2 and c = −3
D=b2-4ac=(2)2-4(8)(-3), where D is discriminant.
=4+96=100>0
Thus, the given equation has distinct and real toots.

Now, using quadratic formula x=-b±D2a, we get
x=-2±10016=-2±1016
=-2+1016,-2-1016=12,-34
Hence, the roots of the given equation are 12 and -34.

(ii) Given -2x2+3x+2=0.
Now, comparing with ax2+bx+c=0, we get
a = −2, b = 3 and c = 2
D=b2-4ac=(3)2-4(-2)(2), where D is discriminant.
=9+16=25>0
Thus, the given equation has distinct and real toots.

Now, using quadratic formula x=-b±D2a, we get
x=-3±252(-2)=-3±5-4
=-3+5-4,-3-5-4=-12,2
Hence, the roots of the given equation are -12 and 2.

(iii) Given 5x2-2x-10=0.
Now, comparing with ax2+bx+c=0, we get
a = 5, b = −2 and c = −10
D=b2-4ac=(-2)2-4(5)(-10), where D is discriminant.
=4+200=204>0
Thus, the given equation has distinct and real toots.

Now, using quadratic formula x=-b±D2a, we get
x=-(-2)±2042×5=2±25110
=1+515,1-515
Hence, the roots of the given equation are 1+515 and 1-515.

(iv) Given 12x-3+1x-5=1, x32,5.
x-5+2x-32x-5x-5=13x-82x2-5x-10x+25=13x-82x2-15x+25=13x-8=2x2-15x+25
2x2-18x+33=0
Now, comparing with ax2+bx+c=0, we get
a = 2, b = −18 and c = 33
D=b2-4ac=(-18)2-4(2)(33), where D is discriminant.
=324-264=60>0
Thus, the given equation has distinct and real toots.

Now, using quadratic formula x=-b±D2a, we get
x=--18±602(2)=18±2154
=18±2154=9+152,9-152
Hence, the roots of the given equation are 9+152 and 9-152.

(v) Given x2+55x-70=0.
Now, comparing with ax2+bx+c=0, we get
a = 1, b = 55 and c = −70
D=b2-4ac=(55)2-4(1)(-70), where D is discriminant.
=125+280=405>0
Thus, the given equation has distinct and real toots.

Now, using quadratic formula x=-b±D2a, we get
x=-55±4052(1)=-55±952
=-55+952,-55-952=25,-75
Hence, the roots of the given equation are 25 and -75.

Page No 42:

Question 2:

Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.

Answer:

Let n be a required natural number.
Square of a natural number diminished by 84 = n2 − 84, and
thrice of 8 more than the natural number = 3(n+8).
Now, by given condition
n2-84=3(n+8)n2-84=3n+24n2-3n-108=0n2-12n+9n-108=0n(n-12)+9(n-12)=0(n-12)(n+9)=0
n=12 or n=-9
Since, -9 is not a natural number, thus n-9.
Hence, the required natural number is 12.

Page No 42:

Question 3:

A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.

Answer:

Let x be required natural number.
Natural number increased by 12 = x + 12, and 160 times its reciprocal = 160x.
Now, by given condition.
x+12=160x
Multiplying both sides by x, we get
x2+12x-160=0x2+(20x-8x)-160=0x2+20x-8x-160=0
x(x+20)-8(x+20)=0(x+20)(x-8)=0

Now, x+20=0x=-20, which is not a natural number, and
 x-8=0x=8.
Hence, the required natural number is 8.

Page No 42:

Question 4:

A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/h more. Find the original speed of the train.

Answer:

Let the original speed of the train =x km/h.
Thus, the increased speed of the train = x+5 km/h and distance = 360 km
Now, by given condition
360x-360x+5=45360(x+5)-360xx(x+5)=45360x+1800-360xx2+5x=45
1800x2+5x=45
x2+5x=1800×54=2250
x2+5x-2250=0x2+50x-45x-2250=0x(x+50)-45(x+50)=0(x+50)(x-45)=0

Now, x+50=0x=-50, which is not possible since speed cannot be negative and 
x-45=0x=45.
Hence, the original speed of the train = 45 km/h

Page No 42:

Question 5:

If Zeba were younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than five times her actual age. What is her age now?

Answer:

Let the actual age of zeba = x years.
Five years ago, her age was  = x-5 years.
Now, by given condition
Square of her age = 11 more than five times her actual age.
∴ x-52=5×x+11
x-52=5x+11x2+25-10x=5x+11x2-15x+14=0x2-14x-x+14=0x(x-14)-1(x-14)
(x-1)(x-14)=0

Now, x=1 not possible because her age five years ago is x-5=1-5=-4 i.e. age cannot be negative
Hence, the required zeba's age now is 14 years.
 

Page No 42:

Question 6:

At present Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age. When Nisha grows to her mother’s present age, Asha’s age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha.

Answer:

Let Nisha's present age be x years.
Then, Asha's present age = x2+2
Now, when Nisha grows to her mother's present age i.e. after x2+2-x years. Then, Asha's age also increased by x2+2-x years.
Again by given condition.
Age of Asha = one year less than 10 times the present age of Nisha.
x2+2+x2+2-x=10x-1
2x2-x+4=10x-12x2-11x+5=02x2-10x-x+5=02xx-5-1x-5=0x-52x-1=0
∴ x=5 or x=12
Here, x=12 is not possible since at x=12, Asha's age is 214 years which is not possible.
Hence, present age of Nisha is 5 years and age of Asha = x2+2=52+2=25+2=27 years.



Page No 43:

Question 7:

In the centre of a rectangular lawn of dimensions 50 m × 40 m, a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be 1184 m2 [see Fig. 4.1]. Find the length and breadth of the pond.

Answer:

Given that a rectangular pond has to be constructed in the centre of a rectangular lawn of dimensions 50 m ⨯ 40 m. So, the distance between pond and lawn would be same around the pond, say x m.



Now, length of rectangular lawn l1=50 m and breadth of rectangular lawn b1=40 m
∴ Length of rectangular pond l2=50-x+x=50-2x and breadth of rectangular pond b2=40-x+x=40-2x
Also, area of the grass surrounding the pond = 1184 m2

∴ Area of rectangular lawn − Area of rectangular pond = Area of grass surrounding the pond.
l1×b1-l2×b2=1184
50×40-50-2x40-2x=11842000-2000-80x-100x+4x2=118480x+100x-4x2=11844x2-180x+1184=0x2-45x+296=0
x2-37x-8x+296=0x(x-37)-8(x-37)=0(x-37)(x-8)=0
∴ x=8 or 37

Now, at x=37, length and breadth of pond are -24 and -34, respectively but length and breadth cannot be negative.
So, x=37 is not possible.

∴ Length of pond = 50-2x=50-2(8)=50-16=34 m.
and breadth of pond = 40-2x=40-2(8)=40-16=24 m.

Hence, required length and breadth of pond are 34 m and 24 m, respectively.

Page No 43:

Question 8:

At t minutes past 2 pm, the time needed by the minutes hand of a clock to show 3 pm was found to be 3 minutes less than t24 minutes. Find t.

Answer:

We know that, the time between 2 pm to 3 pm = 1h = 60 min
Given that, at t min past 2 pm, the time needed by the min hand of a clock to show 3 pm was found to be 3 min less than t24 min i.e.
t+t24-3=60
4t+t2-12=240t2+4t-252=0t2+18t-14t-252=0t(t+18)-14(t+18)=0(t+18)(t-14)=0
∴ t=14 or -18
Since, time cannot be negative, so t-18
Hence, the required value of t is 14 min.



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