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Page No 14.13:

Question 1:

Solving the following quadratic equations by factorization method:
(i) x2+10ix-21=0
(ii) x2+1-2i x-2i=0
(iii) x2-23+3i x+63i=0
(iv) 6x2-17ix-12=0

Answer:

 i x2+10ix-21=0x2+7ix+3ix-21=0xx+7i+3ix+7i=0x+7ix+3i=0x+7i=0 or x+3i=0x=-7i, -3iSo, the roots of the given quadratic equation are -3i and -7i.

ii x2+1-2i x-2i=0 x2+x-2ix-2i=0xx+1-2ix+1=0x+1x-2i=0x+1=0 or x-2i=0x=-1, 2iSo, the roots of the given quadratic equation are -1 and 2i.

iii x2-23+3i x+63i=0 x2-23x-3i x+63i=0xx-23-3ix-23=0x-23x-3i=0x-23=0 or x-3i=0x=23, 3iSo, the roots of the given quadratic equation are 23 and 3i.

iv 6x2-17ix-12=06x2-9ix-8ix-12=03x2x-3i-4i2x-3i=02x-3i3x-4i=02x-3i=0 or 3x-4i=0x=32i, 43iSo, the roots of the given quadratic equation are 32i and 43i.

Page No 14.13:

Question 2:

Solve the following quadratic equations:
(i) x2-32+2i x+62i=0
(ii) x2-5-i x+18+i=0
(iii) 2+i x2-5-i x+2 1-i=0
(iv) x2-2+i x-1-7i=0
(v) i x2-4 x-4i=0
(vi) x2+4ix-4=0
(vii) 2x2+15ix-i=0
(viii) x2-x+1+i=0
(ix) ix2-x+12i=0
(x) x2-32-2i x-2 i=0
(xi) x2-2+i x+2i=0
(xii) 2x2-3+7i x+9i-3=0

Answer:

i x2-32+2i x+62i=0 x2-32 x-2i x+62i=0xx-32 -2ix-32=0x-32x-2i=0x-32=0 or x-2i=0x=32, 2iSo, the roots of the given quadratic equation are 32 and 2i.

ii x2-5-ix+18+i=0Comparing the given equation with the general form ax2+bx+c=0, we geta=1, b=-5-i and c=18+ix=-b±b2-4ac2ax=5-i±5-i2-418+i2x=5-i±5-i2-418+i2x=5-i±-48-14i2x=5-i±i48+14i2x=5-i±i49-1+2×7×i2x=5-i±i7+i22x=5-i±i7+i2x=5-i+i7+i2 or x=5-i-i7+i2x=2+3i, 3-4iSo, the roots of the given quadratic equation are 2+3i and 3-4i.

iii 2+i x2-5-i x+21-i=0Comparing the given equation with the general form ax2+bx+c=0, we geta=2+i, b=-5-i and c=21-ix=-b±b2-4ac2ax=5-i±5-i2-42+i21-i22+ix=5-i±-2i22+ix=5-i±-2i22+i                       ...iLet x+iy=-2i. Then,x+iy2=-2ix2-y2+2ixy=-2i     x2-y2=0 and 2xy=-2              ...iiNow, x2+y22=x2-y22+4x2y2 x2+y22=4x2+y2=2                                         ...iiiFrom ii and iiix=±1 and y=±1As, xy is negative from iix=1, y=-1 or, x=-1, y=1x+iy=1-i or, -1+i-2i=±1-iSubstituting this value in i, we getx=5-i±1-i22+ix=1-i, 45-25iSo, the roots of the given quadratic equation are 1-i and 45-25i.

iv x2-2+i x-1-7i=0Comparing the given equation with the general form ax2+bx+c=0, we geta=1, b=-2+i and c=-1-7ix=-b±b2-4ac2ax=2+i±2+i2+41-7i2x=2+i±7-24i2                                    ... iLet x+iy=7-24i. Then,x+iy2=7-24ix2-y2+2ixy=7-24i     x2-y2=7 and 2xy=-24                          ... iiNow, x2+y22=x2-y22+4x2y2 x2+y22=49+576=625x2+y2=25                                                ... iii            From ii and iiix=±4 and y=±3As, xy is negative From iix=-4, y=3 or, x=4, y=-3x+iy=-4+3i or, 4-3i7-24i=±4-3iSubstituting these values in i, we getx=2+i±4-3i2x=3-i, -1+2iSo, the roots of the given quadratic equation are 3-i and -1+2i.

v ix2-4x-4i=0ix2+4ix-4=0x2+4ix-4=0x+2i2=0x+2i=0x=-2iSo, the roots of the given quadratic equation are -2i and -2i.

vi x2+4ix-4=0x2+2×x×2i+2i2=0x+2i2=0x+2i=0x=-2iSo, the roots of the given quadratic equation are -2i and -2i.

vii 2x2+15 ix-i=0Comparing the given equation with the general form ax2+bx+c=0, we geta=2, b=15 i and c=-ix=-b±b2-4ac2ax=-15 i±15 i2+8i4x=-15 i±8i-154                          ... iLet x+iy=8i-15. Then,x+iy2=8i-15x2-y2+2ixy=8i-15    x2-y2=-15 and 2xy=8                          ... iiNow, x2+y22=x2-y22+4x2y2 x2+y22=225+64=289x2+y2=17                                                ... iii            From ii and iiix=±1and y=±4As, xy is positive       From iix=1, y=4  or, x=-1, y=-4 x+iy=1+4i or, -1-4i8i-15=±1-4iSubstituting these values in i, we get,x=-15 i±1+4i4   x= 1+4-15i4 ,  -1-4+15i4So, the roots of the given quadratic equation are  1+4-15i4 and  -1-4+15i4.

viii x2-x+1+i=0Comparing the given equation with the general form ax2+bx+c=0, we geta=1, b=-1 and c=1+ix=-b±b2-4ac2ax=1 ±1-41+i2x=1±-3-4i2                          ... iLet x+iy=-3-4i. Then,x+iy2=-3-4ix2-y2+2ixy=-3-4i   x2-y2=-3 and 2xy=-4                          ... iiNow, x2+y22=x2-y22+4x2y2 x2+y22=9+16=25x2+y2=5                                              ... iii            From ii and iiix=±1 and y=±2As, xy is negative       From iix=1, y=-2  or, x=-1, y=2x+iy=1-2i or -1+2i-3-4i=±1-2i Substituting these values in i, we getx=1±1-2i 2x=1-i, iSo, the roots of the given quadratic equation are 1-i and i.

ix ix2-x+12i=0ix2+ix+12=0x2+ix+12=0x2+4ix-3ix+12=0xx+4i-3ix+4i=0x+4ix-3i=0x+4i=0 or x-3i=0x=-4i , 3iSo, the roots of the given quadratic equation are -4i and 3i.

x x2-32-2i x-2i=0Comparing the given equation with the general form ax2+bx+c=0, we geta=1, b=-32-2i and c=-2ix=-b±b2-4ac2ax=32-2i±32-2i2+42i2x=32-2i±14-82i2                          ... iLet x+iy=14-82i. Then,x+iy2=14-82ix2-y2+2ixy=14-82i   x2-y2=14 and 2xy=-82                         ... iiNow, x2+y22=x2-y22+4x2y2 x2+y22=196+128=324x2+y2=  18                                                      ... iii            From ii and iiix=±4 and y=±2As, xy is negative       From iix=-4, y=2 or, x=4, y=-2x+iy=4-2 i  or, -4+2 i14-82i=±4-2 iSubstituting these values in i, we getx=32-2i±4-2 i2x=32+4-i2+2 2, 32-4-i2-2 2So, the roots of the given quadratic equation are  32+4-i2+2 2 and 32-4-i2-2 2 .

xi x2-2+i x+2 i=0Comparing the given equation with the general form ax2+bx+c=0, we geta=1, b=-2+i and c=2ix=-b±b2-4ac2ax=2+i±2+i2-42i2x=2+i±1-22 i2      x=2+i±22-12-22 i2x=2+i±2-i22x=2+i±2-i2x=2, i                  So, the roots of the given quadratic equation are  2 and i.

xii 2x2-3+7i x+9i-3=0Comparing the given equation with the general form ax2+bx+c=0, we geta=2, b=-3+7i and c=9i-3x=-b±b2-4ac2ax=3+7i±3+7i2-89i-34x=3+7i±-16-30i4                          ... iLet x+iy=-16-30i. Then,x+iy2=-16-30ix2-y2+2ixy=-16-30i  x2-y2=-16 and 2xy=-30                       ... iiNow, x2+y22=x2-y22+4x2y2 x2+y22=256+900=1156x2+y2=  34                                                     ... iii            From ii and iiix=±3 and y=±5As, xy is negative       From iix=-3, y=5 or, x=3, y=-5x+iy=3-5 i  or, -3+5 i14-82i=±3-5 iSubstituting these values in i, we getx=3+7i±3-5 i4x=3+i2, 3iSo, the roots of the given quadratic equation are  3+i2 and 3i .



Page No 14.15:

Question 1:

Write the number of real roots of the equation (x-1)2+(x-2)2+(x-3)2=0.

Answer:

                 (x -1)2 + (x - 2)2 + (x - 3)2 = 0x2 + 1 - 2x + x2 + 4 - 4x + x2 + 9 - 6x = 0 3x2 - 12 x + 14 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get a = 3 , b=-12 and c = 14.

D=b2-4ac=-122-4×3×14=144-168=-24Since the value of D is less than 0, the given equation has no real roots.

Page No 14.15:

Question 2:

If a and b are roots of the equation x2-px+q=0, than write the value of 1a+1b.

Answer:

Given: x2 - px + q = 0
Also, a and  b are the roots of the given equation.
Sum of the roots = a + b = p        ...(1)
Product of the roots = ab = q       ...(2)

Now, 1a + 1b =  b+aab = pq         [Using equation (1) and (2)]

Hence, the value of 1a + 1b is pq.

Page No 14.15:

Question 3:

If roots α, β of the equation x2-px+16=0 satisfy the relation α2 + β2 = 9, then write the value p.

Answer:

Given equation: x2 - px + 16 = 0
Also,  α and β are the roots of the equation satisfying α2 + β2  = 9.
From the equation, we have:
Sum of the roots = α +β = --p1 = p

Product of the roots = αβ =161 = 16


      Now, α + β2 = α2 + β2 + 2αβ   p2 = 9 + 32 p2  = 41 p =41

Hence, the value of p is 41.     

Page No 14.15:

Question 4:

If 2+3 is root of the equation x2+px+q=0, than write the values of p and q.

Answer:

Irrational roots always occur in conjugate pairs.
If 2+3 is a root and 2-3 is its conjugate root.2+3+2-3=-p4=-9p=-4Also, 2+32-3=q4-3=qq=1



Page No 14.16:

Question 5:

If the difference between the roots of the equation x2+ax+8=0 is 2, write the values of a.

Answer:

Given: x2 + ax + 8 = 0.
Let α and β are the roots of the equation.
 Sum of the roots = α + β =-a1 =-a.
 
  Product of the roots = αβ = 81 = 8
Given: α - β = 2
 
Then, α+ β2 - α - β2 = 4αβ α + β2 - 22 = 4×8 α + β2 - 4 = 32 α + β2 = 32 + 4 = 36α + β = ± 6
 α + β =-a = ±6
∴  a = ±6

Page No 14.16:

Question 6:

Write roots of the equation (a-b)x2+(b-c)x+(c-a)=0.

Answer:

Given:(a-b)x2+(b-c)x+(c-a)=0x2+b-ca-bx+c-aa-b=0x2-c-aa-bx-x+c-aa-b=0                b-ca-b=-c+a-a+ba-b=-c-aa-b-1xx-c-aa-b-1x+c-aa-b=0x-c-aa-bx-1=0x-c-aa-b=0  or  x-1=0x=c-aa-b  or  x=1Thus, roots of the equation are c-aa-b and 1.

Now,
α+β=-b-ca-b1+β=-b-ca-bβ=-b-ca-b-1=c-aa-b

Page No 14.16:

Question 7:

If a and b are roots of the equation x2-x+1=0, then write the value of a2 + b2.

Answer:

Given: x2 - x + 1 = 0
Also, a and b are the roots of the equation.
Then, sum of the roots = a + b = - -11 = 1

Product of the roots = ab = 11 = 1

 a + b2 = a2 + b2 + 2ab 12 = a2 + b2 + 2×1 a2+ b2 = 1 - 2 =-1 a2  + b2 =-1

Page No 14.16:

Question 8:

Write the number of quadratic equations, with real roots, which do not change by squaring their roots.

Answer:

Let α and β be the real roots of the quadratic equation ax2 + bx + c= 0.

On squaring these roots, we get:

α= α2     and       β= β2
α (1 - α) =0  and  β1-β=0 
 α = 0, α = 1  and  β=0, 1

Three cases arise: 

 (i) α = 0, β = 0(ii) α = 1, β = 0(iii) α = 1, β = 1


(i) α = 0, β =0α+β = 0 and αβ = 0

So, the corresponding quadratic equation is,

 x2 - (α+β)x +αβ = 0x2 = 0

(ii) α= 0, β = 1α+β = 1αβ = 0
 
So, the corresponding quadratic equation is,

x2 - (α+β)x +αβ = 0x2 -x+0= 0x2 -x= 0

(iii) α = 1, β = 1α+β = 2αβ =1

So, the corresponding quadratic equation is,

x2 - (α+β)x +αβ = 0x2 -2x+1= 0

Hence, we can construct 3 quadratic equations.

Page No 14.16:

Question 9:

If α, β are roots of the equation x2+lx+m=0, write an equation whose roots are -1αand-1β.

Answer:

Given equation: x2 + lx + m = 0
Also, α and β are the roots of the equation.
Sum of the roots =  α + β = -l1 = -l

Product of the roots = αβ = m1 = m
Now, sum of the roots =  -1α - 1β =-α + βαβ =- -lm = lm
Product of the roots = 1αβ = 1m
 
   x2-Sum of the rootsx+Product of the roots=0x2 - lmx + 1m = 0 mx2 - lx + 1= 0

Hence, this is the equation whose roots are -1α and -1β.

Page No 14.16:

Question 10:

If α, β are roots of the equation x2-a(x+1)-c=0, then write the value of (1 + α) (1 + β).

Answer:

Given: x2 - a(x + 1) - c = 0  or  x2 - ax -a - c = 0 
Also, α and β are the roots of the equation.
 Sum of the roots = α + β = --a1  = a

Product of the roots = αβ = -(a+c)1 = -(a + c)


            (1 + α) (1 + β) = 1 + β + α + αβ                             = 1 + (α + β) + αβ                             = 1 + a - a - c                                     = 1- c

Page No 14.16:

Question 1:

The complete set of values of k, for which the quadratic equation x2-kx+k+2=0 has equal roots, consists of
(a) 2+12
(b) 2±12
(c) 2-12
(d) -2-12

Answer:

(b) 2±12


Since the equation has real roots.D=0b2-4ac=0k2-41k+2=0k2-4k-8=0k=4±16-41-821k=4±2122k=2±12

Page No 14.16:

Question 2:

For the equation  x 2+ x -6=0, the sum of the real roots is
(a) 1
(b) 0
(c) 2
(d) none of these

Answer:

(b) 0

Let p=xp2+p-6=0p2+3p-2p-6=0p+3p-2=0p=-3, 2Also, x=px=2, or x=-3Modulus can not be negative,x=2x=±2x=2 or -2Sum of the roots of x is 0

Page No 14.16:

Question 3:

If a, b are the roots of the equation x2+x+1=0, then a2+b2=
(a) 1
(b) 2
(c) −1
(d) 3

Answer:

(c) −1
Given equation: x2 + x + 1 = 0
Also, a and b are the roots of the given equation.
Sum of the roots = a + b =-Coefficient of xCoefficient of x2=- 11 =-1

Product of the roots = ab = Constant term Coefficient of x2 =11= 1

        (a + b)2 = a2 + b2 + 2ab    (-1)2 = a2 + b2 + 2×1   1 - 2 = a2 + b2 a2 + b2 =-1

Page No 14.16:

Question 4:

If α, β are roots of the equation 4x2+3x+7=0, then 1/α+1/β is equal to
(a) 7/3
(b) −7/3
(c) 3/7
(d) −3/7

Answer:

(d) −3/7

Given equation: 4x2 + 3x + 7 = 0
Also, α and β are the roots of the equation.

Sum of the roots = α + β =-Coefficient of xCoefficient of x2=-34

Product of the roots = αβ =Constant termCoefficient of x2= 74


  ∴     1α + 1β = α + βαβ = -3474 =-37

Page No 14.16:

Question 5:

The values of x satisfying log3 (x2+4x+12)=2 are
(a) 2, −4
(b) 1, −3
(c) −1, 3
(d) −1, −3

Answer:

(d) −1, −3

The given equation is log3(x2+4x+12)=2.

x2+4x+12=32=9x2+4x+3=0x+1x+3=0x=-1, -3

Page No 14.16:

Question 6:

The number of real roots of the equation (x2+2x)2-(x+1)2-55=0 is
(a) 2
(b) 1
(c) 4
(d) none of these

Answer:

(a) 2

x2+2x2-x+12-55=0x2+2x+1-12-x+12-55=0x+12-12-x+12-55=0x+122+1-3x+12-55=0x+122-3x+12-54=0Let p=x+12p2-3p-54=0p2-9p+6p-54=0p+6p-9=0p=9   or    p=-6

Rejecting p=-6x+12=9x2+2x-8=0x2+4x-2x-8=0x+4x-2=0x=2,   x=-4

Page No 14.16:

Question 7:

If α, β are the roots of the equation ax2+bx+c=0, then 1aα+b+1aβ+b=
(a) c / ab
(b) a / bc
(c) b / ac
(d) none of these.

Answer:

(c) b / ac
Given equation: ax2 + bx + c = 0
Also, α and β are the roots of the given equation.

Then, sum of the roots = α + β =-ba
Product of the roots = αβ = ca


  1aα + b + 1aβ + b = aβ + b + aα + b(aα + b) (aβ + b)                                 = a(α + β) + 2ba2αβ + abα + abβ + b2                                 = a(α + β) + 2ba2αβ + abα + β + b2                                =a-ba + 2ba2ca + ab-ba + b2                                = bac 

Page No 14.16:

Question 8:

If α, β are the roots of the equation x2+px+1=0;γ,δ the roots of the equation x2+qx+1=0, then (α-γ)(α+δ)(β-γ)(β+δ)=
(a) q2-p2
(b) p2-q2
(c) p2+q2
(d) none of these.

Answer:

(a) q2-p2
Given: α and β are the roots of the equation x2 + px + 1 = 0.
Also,  γ and δ are the roots of the equation x2 + qx + 1 = 0.
Then, the sum and the product of the roots of the given equation are as follows:
 α + β =-p1 =-pαβ = 11 = 1γ + δ =-q1 =-qγδ = 11 = 1

 
              Moreover, (γ + δ)2 = γ2 + δ2 +2γδγ2 + δ2 = q2 - 2


     (α - γ)  (α + δ) (β - γ) (β + δ) = (α - γ) (β - γ) (α + δ) (β + δ)                                                 =αβ -αγ - βγ + γ2 αβ + αδ + βδ + δ2                                                 =αβ -γα + β + γ2 αβ +δ α + β + δ2                                                 = (1 -γ(-p) + γ2 ) (1 + δ(-p) + δ2 )                                                 = (1 +γp + γ2 ) (1 - δp + δ2 )                                                 =1 -pδ + δ2+ pγ - p2γδ + pγδ2 + γ2- pδγ2 + γ2δ2                                                 =1 -pδ + pγ+ δ2 - p2γδ + pγδ2 + γ2- pδγ2 + γ2δ2                                                 =1 - p(δ - γ) - p2γδ+ pγδ (δ - γ) + (γ2 + δ2) + 1                                                 =1 - p2γδ+ pγδ (δ - γ) - p(δ - γ) + (γ2 + δ2) + 1                                                 =1 - p2+ (δ - γ) p (γδ - 1)+ q2 - 2 + 1                                                 =-p2 + (δ- γ) p (1 - 1) + q2                                                  =q2 - p2

Page No 14.16:

Question 9:

The number of real solutions of 2x-x2-3=1 is
(a) 0
(b) 2
(c) 3
(d) 4

Answer:

(b) 2

Given equation: |2x - x2 - 3| = 1
 
         (i)   2x - x2 - 3 = 1 2x - x2 - 4 = 0x2 - 2x + 4 = 0 (x - 2)2 = 0
          x = 2, 2


(ii)
       -2x + x2 + 3 = 1 x2 - 2x + 2 = 0
     
      x2 - 2x + 1 + 1 = 0 (x - 1)2 - i2 = 0 (x - 1 + i) (x - 1 - i) = 0
      x = 1 - i , 1 + i

Hence, the real solutions are 2, 2.



Page No 14.17:

Question 10:

The number of solutions of x2+x-1=1 is
(a) 0
(b) 1
(c) 2
(d) 3

Answer:

(c) 2
 x2 + |x - 1| = x2+x-1   , x1                      =x2-x+1   , x<1                                 

(i)
x2 + x - 1 = 1x2+x-2=0x2+2x-x-2=0xx+2-1x+2=0x+2x-1=0x+2=0   or, x-1=0x=-2  or x=1

Since -2 does not satisfy the condition x1

(ii)
x2 - x + 1 = 1 x2 - x = 0x2 - x= 0 x ( x - 1) = 0 x= 0   or,    (x - 1) = 0 x = 0,   x = 1

x = 1 does not satisfy the condition x < 1

So, there are two solutions.

Page No 14.17:

Question 11:

If x is real and k=x2-x+1x2+x+1, then
(a) k ∈ [1/3,3]
(b) k ≥ 3
(c) k ≤ 1/3
(d) none of these

Answer:

(a) k ∈ [1/3,3]

k=x2-x+1x2+x+1kx2+kx+k=x2-x+1k-1x2+k+1x+k-1=0

For real values of x, the discriminant of k-1x2+k+1x+k-1=0 should be greater than or equal to zero.

if k1k+12-4k-1k-10 k+12-2k-120k+1+2k-2k+1-2k+203k-1-k+303k-1k-3013k3 i.e. k13, 3-1            ...(i)

And if k=1, then,
x=0, which is real       ...(ii)
So, from (i) and (ii), we get,
k13,3

Page No 14.17:

Question 12:

If the roots of x2-bx+c=0 are two consecutive integers, then b2 − 4 c is
(a) 0
(b) 1
(c) 2
(d) none of these.

Answer:

(b) 1

Given equation: x2 - bx + c = 0
Let α and α+1 be the two consecutive roots of the equation.

Sum of the roots = α + α + 1 = 2α + 1  
Product of the roots = α (α + 1)=α2 + α


 So, sum of the roots= 2α+1=-Coeffecient of xCoeffecient of x2=b1=bProduct of the roots= α2+α=Constant termCoeffecient of x2=c1=cNow, b2-4c=2α+12-4α2+α=4α2+4α+1-4α2-4α=1    

Page No 14.17:

Question 13:

The value of a such that x2-11x+a=0 and x2-14x+2a=0 may have a common root is
(a) 0
(b) 12
(c) 24
(d) 32

Answer:

(a) and (c)

Let α be the common roots of the equations x2-11x+a=0 and x2-14x+2a=0.

Therefore,
   
α2 - 11α + a = 0        ... (1)

α2 - 14α + 2a = 0       ... (2)

Solving (1) and (2) by cross multiplication, we get,

α2-22a+14a=αa-2a=1-14+11α2=-22a+14a-14+11, α=a-2a-14+11α2=-8a-3=8a3, α=-a-3=a3a32=8a3a2= 24aa2-24a=0aa-24=0a=0 or a=24

Disclaimer: The solution given in the book is incomplete. The solution is created according to the question given in the book and both the options are correct.

Page No 14.17:

Question 14:

The values of k for which the quadratic equation kx2+1=kx+3x-11x2 has real and equal roots are
(a) −11, −3
(b) 5, 7
(c) 5, −7
(d) none of these

Answer:

(c) 5, −7

The given equation is kx2+1=kx+3x-11x2 which can be written as.

kx2 + 11x2-kx - 3x+1 =  k+11x2-k+3x+1=0

For equal and real roots, the discriminant of k+11x2-k+3x+1=0.

k+32-4k+11=0k2+2k-35=0k-5k+7=0k=5, -7

Hence, the equation has real and equal roots when k = 5 , -7.

Page No 14.17:

Question 15:

If the equations x2+2x+3λ=0 and 2x2+3x+5λ=0 have a non-zero common roots, then λ =
(a) 1
(b) −1
(c) 3
(d) none of these.

Answer:

(b) −1

Let α be the common roots of the equations, x2 + 2x + 3λ = 0  and  2x2 + 3x + 5λ =0

Therefore,

α2 + 2α + 3λ = 0        ... (1)

2α2 + 3α + 5λ = 0       ... (2)
 
Solving (1) and (2) by cross multiplication, we get

α210λ-9λ=α6λ-5λ=13-4α2=-λ, α=-λ-λ=λ2λ=-1

Page No 14.17:

Question 16:

If one root of the equation x2+px+12=0 is 4, while the equation x2+px+q=0 has equal roots, the value of q is
(a) 49/4
(b) 4/49
(c) 4
(d) none of these

Answer:

(a) 49/4

It is given that, 4 is the root of the equation x2+px+12=0.

16+4p+12=0p=-7

It is also given that, the equation x2+px+q=0 has equal roots. So, the discriminant of
x2+px+q=0 will be zero.

p2-4q=04q=-72=49q=494

Page No 14.17:

Question 17:

The value of p and q (p ≠ 0, q ≠ 0) for which p, q are the roots of the equation x2+px+q=0 are
(a) p = 1, q = −2
(b) p = −1, q = −2
(c) p = −1, q = 2
(d) p = 1, q = 2

Answer:

(a) p = 1, q = −2

It is given that, p and q (p ≠ 0, q ≠ 0) are the roots of the equation x2+px+q=0.

Sum of roots=p+q=-p2p+q=0          ... (1)

Product of roots=pq=qqp-1=0p=1, q=0 but q0

Now, substituting p = 1 in (1), we get,

2+q=0q=-2

Disclaimer: The solution given in the book is incorrect. The solution here is created according to the question given in the book.

Page No 14.17:

Question 18:

The set of all values of m for which both the roots of the equation x2-(m+1)x+m+4=0 are real and negative, is
(a) (-,-3][5,)
(b) [−3, 5]
(c) (−4, −3]
(d) (−3, −1]

Answer:

(c) m(-4,-3]

The roots of the quadratic equation x2-(m+1)x+m+4=0 will be real, if its discriminant is greater than or equal to zero.

m+12-4m+40m-5m+30m-3 or m5    ... (1)

It is also given that, the roots of x2-(m+1)x+m+4=0 are negative.
So, the sum of the roots will be negative.

Sum of the roots < 0

m+1<0m<-1                ... (2)

and product of zeros >0

m+4>0m>-4                  ...(3)

From (1), (2) and (3), we get,

m(-4,-3]

Disclaimer: The solution given in the book is incorrect. The solution here is created according to the question given in the book.

Page No 14.17:

Question 19:

The number of roots of the equation (x+2)(x-5)(x-3)(x+6)=x-2x+4 is
(a) 0
(b) 1
(c) 2
(d) 3

Answer:

(b) 1

(x +2) (x - 5)(x -3) (x + 6) = (x - 2)(x + 4) (x2 - 3x - 10) (x + 4) = (x2 + 3x - 18) (x - 2) x3 + 4x2  - 3x2 - 12x - 10x - 40 = x3 - 2x2 + 3x2 - 6x -18x + 36x2 - 22x - 40 = x2 - 24x + 36 2x = 76 x = 38
Hence, the equation has only 1 root.

Page No 14.17:

Question 20:

If α and β are the roots of 4x2+3x+7=0, then the value of 1α+1β is
(a) 47
(b) -37
(c) 37
(d) -34

Answer:

(b) −3/7

Given equation: 4x2 + 3x + 7 = 0
Also, α and β are the roots of the equation.

Then, sum of the roots = α + β =-Coefficient of xCoefficient of x2=-34

Product of the roots = αβ =Constant termCoefficient of x2= 74

  1α + 1β = α + βαβ = -3474 =-37

Page No 14.17:

Question 21:

If α, β are the roots of the equation x2+px+q=0 then -1α+1β are the roots of the equation
(a) x2-px+q=0
(b) x2+px+q=0
(c) qx2+px+1=0
(d) qx2-px+1=0

Answer:

(d) qx2-px+1=0
Given equation: x2 + px + q = 0
Also, α and β are the roots of the given equation.
Then, sum of the roots = α + β =-p
Product of the roots = αβ = q
Now, for roots -1α , -1β, we have:
       Sum of the roots = -1α - 1β = - α + βαβ =--pq = pq
      Product of the roots = 1αβ = 1q
Hence, the equation involving the roots -1α,-1β is as follows:
x2-α+βx+αβ=0

  x2 - pqx + 1q = 0 qx2 -px + 1 = 0

Page No 14.17:

Question 22:

If the difference of the roots of x2-px+q=0 is unity, then
(a) p2+4q=1
(b) p2-4q=1
(c) p2+4q2=(1+2q)2
(d) 4p2+q2=(1+2p)2

Answer:

(b) p2-4q=1
Given equation: x2 - px + q = 0
Also α and β are the roots of the equation such that α - β = 1.
Sum of the roots = α + β =-Coefficient of xCoefficient of x2= --p1 = p

Product of the roots = αβ =Constant termCoefficient of x2= q

    (α + β)2 - (α - β)2 = 4αβp2 - 1 = 4qp2 - 4q = 1



Page No 14.18:

Question 23:

If α, β are the roots of the equation x2-p(x+1)-c=0, then (α+1)(β+1)=
(a) c
(b) c − 1
(c) 1 − c
(d) none of these

Answer:

(c) 1 − c
Given equation: x2 - p(x + 1) - c = 0 or    x2 - px-p- c=0 
Also  α and β are the roots of the equation.
Sum of the roots = α + β = p
Product of the roots = αβ = -(c + p)
 

Then, (α + 1) (β + 1) = αβ + α + β + 1                                        = -(c + p) + p + 1                                        = -c - p + p + 1                                      = 1- c

Page No 14.18:

Question 24:

The least value of k which makes the roots of the equation x2+5x+k=0 imaginary is
(a) 4
(b) 5
(c) 6
(d) 7

Answer:

(d) 7

The roots of the quadratic equation x2+5x+k=0 will be imaginary if its discriminant is less than zero.

25-4k<0k>254

Thus, the minimum integral value of k for which the roots are imaginary is 7.

Page No 14.18:

Question 25:

The equation of the smallest degree with real coefficients having 1 + i as one of the roots is
(a) x2+x+1=0
(b) x2-2x+2=0
(c) x2+2x+2=0
(d) x2+2x-2=0

Answer:

(b) x2-2x+2=0

We know that, imaginary roots of a quadratic equation occur in conjugate pair.

It is given that, 1 + i is one of the roots.

So, the other root will be 1-i.

Thus, the quadratic equation having roots 1 + i and 1 - i is,

x2-1+i+1-ix+1+i1-i=0x2-2x+2=0



Page No 14.5:

Question 1:

x2 + 1 = 0

Answer:

Given: x2 + 1 = 0
    x2 + 1 = 0 x2 - 1i2 = 0 (x + i) (x - i) = 0                       [ ( a2 - b2) = (a + b) (a - b)]
 (x + i) = 0 or (x - i) = 0
 x =-i   or  x = i
Hence, the roots of the equation are i  and -i.

Page No 14.5:

Question 2:

9x2 + 4 = 0

Answer:

Given: 9x2 + 4 = 0
 
    9x2 + 4 = 0 (3x)2 + 22 = 0 (3x)2 - (2i)2 = 0 (3x + 2i) (3x - 2i) = 0              [(a2 - b2) = (a + b) (a - b)]
 (3x + 2i) = 0   or,  (3x - 2i) = 0
 3x =-2i  or   3x = 2i
 x = -2i3    or  x = 2i3
Hence, the roots of the equation are 2i3 and -2i3.

Page No 14.5:

Question 3:

x2 + 2x + 5 = 0

Answer:

Given: x2 + 2x + 5 = 0

    x2 + 2x + 5 = 0x2 + 2x + 1 + 4 = 0 (x + 1)2 - (2i)2 = 0                [(a + b)2 = a2 + b2 + 2ab] (x + 1 + 2i) (x + 1 - 2i)  = 0    [ a2 - b2 = (a + b) (a - b)]
 (x + 1 + 2i) = 0 or,  (x + 1 - 2i) = 0
 x = -(1+ 2i)   or,    x = -1 + 2i
Hence, the roots of the equation are -1 + 2i  and -1 - 2i.    

Page No 14.5:

Question 4:

4x2 − 12x + 25 = 0

Answer:

We have:
4x2 - 12x + 25 = 0 4x2 - 12 x + 9 + 16 = 0 (2x)2 + 32 - 2×2x×3 - (4i)2 = 0(2x- 3)2 - (4i)2 = 0 (2x - 3 + 4i) (2x - 3 -4i) = 0       [a2 - b2 = (a + b) (a - b)]
 (2x - 3 + 4i) = 0  or,  (2x - 3 - 4i) = 0
 2x = 3 - 4i   or,     2x = 3 + 4i
x = 32 - 2i  or,  x = 32 + 2i
Hence, the roots of the equation are 32 - 2i and  32 + 2i.  
    

Page No 14.5:

Question 5:

x2 + x + 1 = 0

Answer:

We have:
                x2 + x + 1 = 0 x2 + x + 14 + 34 = 0 x2 +  122 +2×x×12- 3i22 = 0x + 122 - 3i22 = 0 x + 12 + 3i2 x + 12 - 3i2 = 0
 
            x + 12 + 3i2 = 0  or,  x + 12 - 3i2 = 0
           
            x = -12 - 3i2   or,  x =-12 + 3i2

  Hence, the roots of the equation are -12 - i32 and -12 + i32.        



Page No 14.6:

Question 6:

4x2+1=0

Answer:

We have:
                  4x2 + 1 = 0 (2x)2 - i2 = 0 (2x)2 - (i)2 = 0 (2x + i) (2x - i) = 0
   
                (2x + i) = 0  or  (2x - i) = 0
                2x = -i    or   2x = i
                 x =-i2   or  x = i2
Hence, the roots of the equation are 12i  and -12i.

Page No 14.6:

Question 7:

x2-4x+7=0

Answer:

We have:
                   x2 - 4x + 7 = 0 x2 - 4x + 4 + 3 = 0 x2 - 2×x×2 + 22 - (3i)2 = 0 (x - 2)2 - (3i)2 = 0 (x - 2 + 3i) (x - 2 - 3i) = 0
    
                (x - 2 + 3i) = 0   or,   (x - 2 - 3i) = 0
                x = 2 - 3i       or,        x = 2 + 3i

Hence, the roots of the equation are 2 ± i3.

Page No 14.6:

Question 8:

x2+2x+5=0

Answer:

We have:
              x2 + 2x + 2 = 0 x2 + 2x + 1 + 1 = 0 x2 + 2×x×1 + 12 - (i)2 = 0 (x + 1)2 - (i)2 = 0 (x + 1 + i) (x + 1 - i) = 0
 
               (x + 1 + i) = 0   or   (x + 1 - i) = 0
               x = -1 - i   or   x = -1 + i
         
 Hence, the roots of the equation are -1 + i and -1 - i.

Page No 14.6:

Question 9:

5x2-6x+2=0

Answer:

Given:  5x2 - 6x + 2 = 0
  Comparing the given equation with general form of the quadratic equation ax2+ bx + c = 0, we get a = 5, b =-6 and c = 2.

Substituting these values in α = -b + b2 - 4ac2a and β = -b - b2 - 4ac2a, we get:

  α = 6 + 36 - 4×5× 22×5   and    β= 6 - 36 - 4×2×52×5

 α = 6 + -410            and          β =  6 - -410

 α = 6 + 4i210        and       β= 6 - 4i210
 
 α = 6 + 2i10       and          β = 6 - 2i10

α = 2 ( 3 + i)10   and         β =  2 ( 3 - i)10

 α = 35 + 15i      and       β = 35 - 15i
    

Hence, the roots of the equation are 35 ± 15i.      

Page No 14.6:

Question 10:

21x2+9x+1=0

Answer:

Given:  21x2 + 9x + 1 = 0
Comparing the given equation with  the general form of the quadratic equation ax2 + bx + c = 0, we get a = 21, b = 9 and c = 1.


Substituting these values in α = -b + b2 - 4ac2a  and β= -b - b2 - 4ac2a, we get:

     α = -9 + 81 - 4×21×12×21      and      β= -9 - 81 - 4×21×12×21

 α = -9 + 3i42        and       β = -9 - 3i42

α = -942 + 3i42      and       β=- 942 - 3i42

 α = -314 + 3i42    and       β=-314 - 3i42


Hence, the roots of the equation are -314 ± i342.

Page No 14.6:

Question 11:

x2-x+1=0

Answer:

We have:
      x2 - x + 1 = 0 x2 - x + 12 + 34 = 0x2 -2× x ×12+ 122 - 34i2 = 0 x - 122 - i322 =0 x - 12 + i32 x - 12 - i32 = 0
 
x - 12 + i32 = 0    or     x - 12 - i32 = 0
  
x = 12 - i32      or    x = 12 + i32
Hence, the roots of the equation are 12 ± i32.

Page No 14.6:

Question 12:

x2+x+1=0

Answer:

We have:
    x2 + x + 1 = 0 x2 + x + 14 + 34 = 0 x + 122 - 34i2 = 0 x + 12 2 - i322 = 0 x + 12 + i32  x + 12 - i32 = 0

 x + 12 + i32 = 0    or    x + 12 - i32 = 0
      
 x =-12 - i32     or      x =-12 + i32

 Hence, the roots of the equation are -12 ± i32.

Page No 14.6:

Question 13:

17x2-8x+1=0

Answer:

Given:    17x2 - 8x + 1 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get a = 17, b =-8  and c = 1.


Substituting these values in   α= -b + b2 - 4ac2a and  β = -b - b2 - 4ac2a, we get:

        α=8 + 64 - 4×17×12×17      and      β = 8 - 64 - 4×17 ×12×17

   α = 8 + 64 - 6834         and         β = 8 - 64 - 6834

  α = 8 + -434           and          β = 8 - -434

  α = 8 + 4i234    and      β= 8 - 4i234

 α = 8 + 2i34        and        β = 8 - 2i34

  α = 4 + i17       and    β= 4 - i17

 α = 417 + 117i    and    β= 417 - 117i

Hence, the roots of the equation are 417 ± 117i.

Page No 14.6:

Question 14:

27x2-10+1=0

Answer:

Given: 27x2 - 10x + 1 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get a = 27, b =-10 and  c = 1.

Substituting these values in α = -b + b2 - 4ac2a and β = -b - b2 - 4ac2a, we get:

              α= 10 + 100 - 4×27×12×27       and      β = 10 - 100 - 4×27×12×27

         α = 10 + 100 - 10854        and      β = 10 - 100 - 10854
      
        α = 10 + -854        and        β = 10 - -854
     
        α = 10 + 8i254     and       β = 10 - 8i254
  
       α = 10 + i2254      and    β= 10 - i2254
    
      α = 2(5 + i2) 54      and      β = 2(5 - i2)54
    
     α = 527 + 227i      and       β= 527 - 227i


Hence, the roots of the equation are 527 ± 227i.

Page No 14.6:

Question 15:

17x2+28x+12=0

Answer:

Given: 17x2 + 28x + 12 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get a = 17, b = 28  and c= 12.


Substituting these values in  α = -b + b2 - 4ac2a and β = -b - b2- 4ac2a, we get:

         α = -28 + 784 - 4×17×1234       and       β = -28 - 784 - 4×17×1234

    α = -28 + 784 - 81634      and       β = -28 - 784 - 81634
  
   α = -28 + -3234       and         β = -28 - -3234

  α = -28 + 32i234      and        β= -28 - 32i234

  α = -28 + 42 i34       and      β = -28 - 42 i34
 
  α = -14 + 22 i17      and       β = -14 - 22 i17


Hence, the roots of the equation are  -1417 ± 2217i.

Page No 14.6:

Question 16:

21x2-28x+10=0

Answer:

Given:  21x2 - 28x + 10 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get a = 21, b =-28  and   c = 10.


Substituting these values in α = -b + b2 - 4ac2a and  β = -b - b2 - 4ac2a, we get:

           α = 28 + 784 - 4×21×102×21    and    β = 28 - 784 - 4×21×102×21

      α = 28 + -5642     and      β = 28 - -5642
  
     α = 28 +2i1442   and    β = 28 - 2i1442

     α = 14 + i1421    and    β = 14 - i1421

    α = 23 + 1421i      and    β = 23 - 1421i



Hence, the roots of the equation are 23 ± 1421.

Page No 14.6:

Question 17:

8x2-9x+3=0

Answer:

Given: 8x2 - 9x + 3 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get  a = 8, b=-9 and  c = 3.


Substituting these values in  α= -b + b2- 4ac2a  and β = -b - b2 - 4ac2a, we get:

                       α = 9 + 81 - 4×8×32×8    and      β= 9 - 81 - 4×8×32×8

                  α = 9 + 81 - 9616         and         β = 9 - 81 - 9616

                  α = 9 + -1516         and           β= 9 - -1516

                  α = 9 + 15i216      and       β = 9 - 15i216

                   α = 9 + i1516    and        β = 9 - i1516

                  α = 916 - 1516i    and       β = 916 + 1516i

Hence, the roots of the equation are 916 ± 1516i.

                 

Page No 14.6:

Question 18:

13x2+7x+1=0

Answer:

Given:  13x2 + 7x + 1 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get  a = 13, b = 7  and c = 1.

Substituting these values in  α = -b + b2 - 4ac2a  and  β= -b - b2 - 4ac2a, we get:

                  α = -7 + 49 - 4×13×12×13     and         β= -7 - 49 - 4×13×12×13
    
             α = -7 + 49 - 5226         and          β = -7 - 49 - 5226

             α = -7 + -326           and          β = -7 - -326

            α = -7 + i326       and         β = -7 - i326

          α = -726 + 326i    and        β = -726 - 326i


Hence, the roots of the equation are  -726 ± 326i.

Page No 14.6:

Question 19:

2x2+x+1=0

Answer:

Given:   2x2 + x +1 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get a = 2, b = 1 and  c = 1.

Substituting these values in  α = -b + b2 - 4ac2a and  β = -b - b2 - 4ac2a, we get:
                         
                     α = -1 + 1 - 4×2×12×2        and          β = -1 - 1 - 4×2×12×2

                  α = -1 + -74                 and                β = -1 - -74

                 α = -1 + i74           and         β = -1 - i74

                 α =-14 + 74i        and         β =-14 - 74i


Hence, the roots of the equation are -1 ± i74.

Page No 14.6:

Question 20:

3x2-2x+33=0

Answer:

Given: 3x2 - 2x + 33 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get  a = 3, b = -2  and   c = 33.


Substituting these values in  α = -b + b2 - 4ac2a and  β = -b - b2 - 4ac2a, we get:

                α = 2 + 2 - 4×3×3323       and         β = 2 - 2 - 4×3×3323

           α =  2 + -3423             and               β = 2 - -3423

           α = 2 + i3423         and             β = 2 - i3423

Hence, the roots of the equation are 2 ± i3423.

Page No 14.6:

Question 21:

2x2+x+2=0

Answer:

Given: 2x2 + x + 2 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get a = 2, b = 1 and c =2.


Substituting these values in α = -b + b2 -4ac2a and  β = -b - b2 - 4ac2a, we get:
 
                     α = -1 + 1 - 4×2×222    and     β = -1 - 1 - 4×2×222

                 α =-1 + -722          and            β = -1 - -722
  
                α = -1 + i722      and         β = -1 - i722

Hence, the roots of the equation are -1 ± i722.
            

Page No 14.6:

Question 22:

x2+x+12=0

Answer:

Given equation: x2 + x + 12 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get a = 1, b = 1 and c = 12.

Substituting these values in α = -b + b2 - 4ac2a and β = -b - b2 - 4ac2a, we get:

     α = -1 + 1 - 4×122   and        β = -1 - 1 - 4×122

α = -1 + 1 - 222     and     β = -1 - 1 - 222

 α = -1 + i22 - 12  and     β =-1 - i22 - 12
 

Hence, the roots of the equation are -1 ± i22 - 12.

Page No 14.6:

Question 23:

x2+x2+1=0

Answer:

Given equation: x2 + x2 + 1 = 0
Comparing the given equation with  the general form of the quadratic equation ax2 + bx + c = 0, we get a = 1, b= 12 and c = 1.

Substituting these values in  α = -b + b2 - 4ac2a  and  β = -b - b2 - 4ac2a, we get:

      α = -12 + 12- 4×1×12          and              β = -12 - 12 - 4×1×12

     α = -12 + -722                 and                   β = -12 - -722

    α = -12 + i722                  and                        β = -12 - i722
 
  α = -1 + i722                     and                         β = -1 - i722


Hence, the roots of the equation are -1 ± i722.

Page No 14.6:

Question 24:

5x2+x+5=0

Answer:

Given: 5x2 + x + 5 = 0  
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get a = 5 , b = 1 and  c = 5.


Substituting these values in α= -b + b2 - 4ac2a  and  β = -b - b2 - 4ac2a, we get:

  α = -1 + 1 - 4×5 × 525          and               β = -1 - 1 - 4×5 × 525

 α = -1 + -1925                     and                         β = -1 - -1925

 α = -1 + i1925                      and                         β = -1 - i1925



Hence, the roots of the equation are -1 ± i1925.

Page No 14.6:

Question 25:

-x2+x-2=0

Answer:

  -x2 +x- 2 = 0  x2  - x + 2 = 0 x2 - x + 14 + 74 = 0 x2 -2× x×12 + 122 - 74i2 = 0 x - 122 - i722 = 0 x - 12 + i72 x - 12 - i72 = 0

  x - 12 + 72i = 0  or,    x - 12 - 72i = 0
   
x = 12 - 72i   or,    x = 12 + 72i


Hence, the roots of the equation are 12 ± 72i.

Page No 14.6:

Question 26:

x2-2x+32=0

Answer:

     x2 - 2x + 32 = 0 x2  - 2x + 1 + 12 = 0 x -12 - 12i2 = 0 x - 1 + 12i x - 1 - 12i = 0

 x - 1 - 12i = 0  or,    x - 1 +12i = 0

 x = 1 + 12i      or,         x = 1 - 12i

Hence, the roots of the equation are  1 ± 12i.

Page No 14.6:

Question 27:

3x2-4x+203=0

Answer:

Given:   3x2 - 4 x + 203 = 0     
Comparing the given equation with the general form of the quadratic equation ax2 + bx = c = 0, we get a = 3, b=-4 and c = 203.


Substituting these values in α = -b + b2 - 4ac2a and  β = -b - b2 - 4ac2a, we get:

  α = 4 + 16 - 4×3×2036       and          β = 4 - 16 - 4×3×2036

 α = 4+ -646   and      β = 4 - -646

 α = 4 + 8i6        and      β = 4 - 8i6

α = 2 + 4i3       and        β= 2 - 4i3


Hence, the roots of the equation are  2 ± 4i3.



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