Rd Sharma Xi 2020 _volume 1 Solutions for Class 11 Science Math Chapter 14 Quadratic Equations are provided here with simple step-by-step explanations. These solutions for Quadratic Equations are extremely popular among Class 11 Science students for Math Quadratic Equations Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi 2020 _volume 1 Book of Class 11 Science Math Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi 2020 _volume 1 Solutions. All Rd Sharma Xi 2020 _volume 1 Solutions for class Class 11 Science Math are prepared by experts and are 100% accurate.

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:

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 14.15:

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 14.16:

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 14.16:

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 14.16:

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 14.16:

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 14.16:

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 14.16:

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 14.16:

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 14.16:

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 14.17:

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 14.17:

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 14.17:

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 14.17:

Question 1:

If 1 – i is a root of the equation x2 + ax + b = 0, where a, bR, then the values of a and b are ____________.

Answer:

for equation x2+ax+b=0 one root is 1-i.Let us suppose other root is p+iq when p, qSum of roots of quadratic equations is=-ai.e. 1-i+p+iq=-ai.e. 1+p+i (-1+q)=-asince a is real given-1+q=0            i.e. imaginary part is zeroi.e. q=1also product of roots=b1-ip+iq=bi.e. p-i2q-ip+iq=bi.e. p+q+iq-p=bSince b is also realp-q=0p=q=1Hence a=1+p=-2, b=p+q=1+1=2

Page No 14.17:

Question 2:

If the difference of the roots of the equation x2Px + 8 = 0 is 2, then P =___________.

Answer:

For x2-Px+8=0Let us suppose two roots are z1 and z2given: z1-z2=2also z1z2=8                     product of rootsi.e. z2+2Z2=8i.e. z22+2z2=8i.e. z22+2z2-8=0i.e. z22+4z2-2z2-8=0i.e. z2+4 z2-2=0i.e. z2=-4 or z2=2z1=-2 or z2=4Since P=z1+z2P=-4-2 or P=4+2i.e. P=-6 or 6

Page No 14.17:

Question 3:

If the equation 2x2kx + x + 8 = 0 has real and equal roots, then k = __________.

Answer:

for 2x2-kx+x+8=0given; roots one real and equali.e. D=0          Discriminati.e. b2-4ac=0Here b=-k+1a=2c=8with reference to standard equationsax2+bx+c=01-K2-428=0i.e. K2+1-2K-64=0i.e. K2-2K-63=0i.e. K2-9K+7K-63=0i.e. K-9k+7=0i.e. K=9 or -7

Page No 14.17:

Question 4:

The number of real roots of the equation x2 + 5 |x| + 4 = 0 is _________.

Answer:

x2+5 x+4=0for x, we have 2 casescase iif x>0Quadratic equation is x2+5x+4=0i.e. x2+4x+x+4=0i.e. x+4 x+1=0i.e. x=-1 or -4which is not possible since x>0caseii for x<0,Quadratic equation is x2-5x+4=0i.e. x2-4x-x+4=0i.e. x-4 x-1=0i.e. x=1 or x=4which is not possible, since x is negative in this caseHence, no real solution exist for x2+5x+4=0



Page No 14.18:

Question 5:

If one root of the equation x2 + px + 12 = 0 is 4, then the sum of the roots is ____________.

Answer:

One root of x2+px+12=0 is 4Let us suppose the other solution is z. Sum of roots=-p1=-pi.e. z+4=-pand product of roots is 12i.e. 4z=12i.e. z=3sum of roots is 4+3=7

Page No 14.18:

Question 6:

If α, β are roots of the equation x2 + x + 1 = 0, then the equation whose roots are α19 and β7 is ____________.

Answer:

Given : α and β are roots of x2+x+1=0i.e. α2+α+1=0 and β2+β+1=0Since x3-1=x-1 x2+x+1α3-1=α-1 α2+α+1and β3-1=β-1 β2+β+1α3-1=0 and β3-1=0i.e. α3=1 and β3=1Now, α19+β7=(α3)6·α+(β3)2β=1·α+1·β   α3=1 and β3=1
=α+β=-1 α and β are roots of x2+x+1i.e. α19+ β7=-1and α19 β7=α36αβ32β                   = αβ=1               α and β are roots of x2+x+1Hence α19 and β7 satisfies quadratic equation x2--1x+1=0i.e. x2+x+1=0

Page No 14.18:

Question 7:

The value of 6+6+6+......to  is ____________.

Answer:

Let 6+6+6+_____=x=6+x=xi.e. 6+x=x2               squaring both sidesi.e. x2-x-6=0i.e. x2-3x+2x-6=0i.e. x-3 x+2=0i.e. x=3 or x=-2Since x=-2 is not possible                   (being negative)x=3i.e. 6+6+6+_____=3

Page No 14.18:

Question 8:

If the equations px2 + 2qx + r = 0 and qx2-2prx+q=0 have real roots, then q2 =____________.

Answer:

Given px2+2qx+r=0and qx2-2pr x+q=0 have real rootsfor px2+2qx+r=0Since roots are realDiscriminant D0i.e. b2-4ac0i.e. 2q2-4pr0i.e. 4q2-4pr0i.e. q2pr           ...ialso for qx2-2pr x+q=0D0 i.e. b2-4ac0i.e. 4pr-4qq0i.e. prq2i.e. q2pr              ...ii from i and iiprq2prq2=pr

Page No 14.18:

Question 9:

If the roots of the equation x2 – 8x + a2 – 6a = 0 are real, then 'a' lies in the interval ____________.

Answer:

For x2-8x+a2-6a=0roots are realHere a=1, b=-8, c=a2-6ai.e. D0       (Discriminant)i.e. b2-4ac0i.e. 64-41a2-6a0i.e. 64-4a2+24a0i.e. -4a2+24a+640Now, multiplying by -14i.e. 4a24-24a4-6440i.e. a2-6a-160i.e. a2-8a+2a-160i.e. aa-8+2a-80i.e. a+2 a-80

for a < –2, (a + 2) (a – 8) ≥ 0 and for a > 8, (a + 2) (a – 8) ≥ 0
 

Page No 14.18:

Question 10:

If the equations x2 + x + a = 0 and x2 + ax + 1 = 0, a ≠ 1, have a common root, then a = ____________.

Answer:

Given :x2+x+a=0 andx2+ax+1=0 have a common root where a1.Solving above 2 equations,we getx2+x+a=0x2+ax+1=0---x1-a+a-1=0i.e. x1-a=+1-ai.e. x=1            a1 given a can be obtained form x2+x+a=0Since x=1 satisfy x2+x+a=0 i.e. 12+1+a=0i.e. a=-2

Page No 14.18:

Question 11:

If the quadratic equation 2x2 – (a3 + 8a – 1) x + a2 – 4a = 0 possesses roots of opposite signs, then a lies in the interval ____________.

Answer:

For 2x2a3+8a1 x+a24a=0 has roots of opposite sign.Let α and β be 2 roots of above equation.Then, according to give condition,αβ<0also, since sum of roots=a3+8a-12α+β=a3+8a-12and product of roots is a2-4ai.e. αβ=a2-4asince αβ<0a2-4a<0i.e. aa-4<0

Since a(a – 4) < 0 is true for 0 < a < 4
Hence a ∈ (0, 4)

Page No 14.18:

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 14.18:

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 14.18:

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 14.18:

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 14.18:

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 14.18:

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 14.18:

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 14.18:

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 14.18:

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 14.18:

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 14.5:

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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