# Class 10 RD Sharma Solutions – Chapter 8 Quadratic Equations – Exercise 8.6 | Set 1

### Question 1. Determine the nature of the roots of following quadratic equations :

(i) 2xÂ² â€“ 3x + 5 = 0

(ii) 2xÂ² â€“ 6x + 3 = 0

(iii) 3/5 xÂ² â€“ 2/3 x + 1 = 0

(iv) 3xÂ² â€“ 4âˆš3 x + 4 = 0

(v) 3xÂ² â€“ 2âˆš6 x + 2 = 0

Solution:

(i) 2xÂ² â€“ 3x + 5 = 0

Here a=2, b=-3, c=5

D=b2-4ac=(-3)2-4*2*5

=-9-40=-31

D<0

Roots are not real

(ii) 2xÂ² â€“ 6x + 3 = 0

Here a=2, b=-6, c=3

D=b2-4ac

=(-6)2-4*2*3=36-24=12

D>0

Roots are real and distinct

(iii) 3/5 xÂ² â€“ 2/3 x + 1 = 0

Here a=3/5, b=-2/3, c=1

Discriminant (D)=b2-4ac

D<0

Roots are not real

(iv) 3xÂ² â€“ 4âˆš3 x + 4 = 0

Here a=3, b=-4âˆš3, c=4

D=b2-4ac

=(-4âˆš3)2-4*3*4=48-48=0

D=0

Roots are real and equal

(v) 3xÂ² â€“ 2âˆš6 x + 2 = 0

Here a=3, b=â€“ 2âˆš6, c=2

Discriminant (D)=b2-4ac

=(â€“ 2âˆš6)2-4*3*2=24-24=0

D=0

Roots are real and equal

### Question 2. Find the values of k for which the roots are real and equal in each of the following equations :

(i) kxÂ² + 4x + 1 = 0

Solution:

Here a=k, b=4, c=1

Discriminant(D)=b2-4ac

=(4)2-4*k*1

=16-4k

Roots are real and equal

D=0

16-4k=0â‡’4k=16

k=16/4=4

Hence k=4

(ii) kxÂ² â€“ 2âˆš5 x + 4 = 0

Solution:

Here a=k, b=-2âˆš5, c=4

Discriminant(D)=b2-4ac

=( â€“ 2âˆš5 )-4*k*4=20-16k

Roots are real and equal

D=0

20-16k=0â‡’ 16k=20

k=20/16=5/4

Hence k=5/4

(iii) 3xÂ² â€“ 5x + 2k = 0

Solution:

Here a=3, b=-5, c=2k

Discriminant (D)=b2-4ac

=(-5)2-4*3*2k

=25-24k

Roots are real and equal

D=0

25-24k=0â‡’24k=25

k=25/24

(iv) 4xÂ²+ kx + 9 = 0

Solution:

Here a=4, b=k, c=9

Discriminant (D)=b2-4ac

=k2-4*4*9=k2-144

Roots are real and equal

k2-144=0â‡’k2=144=(Â±12)2

(v) 2kxÂ² â€“ 40x + 25 = 0

Solution:

Here a=2k, b=-40, c=25

Discriminant(D)=b2-4ac

=(-40)2-4*2k*25

=1600-200k

Roots are real and equal

D=0

1600-200k=0â‡’200k=1600

k=1600/200=8

Hence k=8

(vi) 9xÂ² â€“ 24x + k = 0

Solution:

Here a=9, b=-24, c=k

Discriminant(D)=b2-4ac

=(-24)2-4*9*k

=576-36k

Roots are real and equal

D=0

576-36k=0

36k=576â‡’k=576/36=16

k=16

(vii) 4xÂ² â€“ 3kx +1 = 0

Solution:

Here a=4, b=-3k, c=1

Discriminant (D)=b2-4ac

=(-3k)2-4*4*1

=9k2-16

Roots are real and equal

D=0

9k2-16=0â‡’9k2=16

k2=16/9=

(viii) xÂ² â€“ 2 (5 + 2k) x + 3 (7 + 10k) = 0

Solution:

Here a=1, b=-2(5+2k) and c=3(7+10k)

Discriminant (D)=b2-4ac

[-2(5+2k)]2-4*1*3(7+10k)

=4(25+4k2+20k)-12(7+10k)

=100+16k2+80k-84-120k

16k2-40k+16

Roots are real and equal

D=0

16k2-40k+16=0

2k2-5k+2=0

2k2-4k-k+2=0

2k(k-2)-1(k-2)=0

(k-2)(2k-1)=0

Either k-2=0, then k=2 or 2k-1=0, then 2k=1â‡’k=1/2

Hence, k=2, 1/2

(ix) (3k + 1) xÂ² + 2(k + 1) x + k = 0

Solution:

Here a=3k+1, b=2(k+1), c=k

Discriminant(D)=b2-4ac

=[2(k+1)]2-4*(3k+1)*k

=4(k2+2k+1)-4k(3k+1)

=4k2+8k+4-12k2-4k

-8k2+4k+4

Roots are real and equal

D=0

-8k2+4k+4=0

2k2-k-1=0 (Dividing by -4)

2k2-2k+k-1= {Therefore -2=-2*1

-1=-2+1}

2k(k-1)+1(k-1)=0

(k-1)(2k+1)=0

Either k-1=0, then k=1 or 2k+1=0, then 2k=-1â‡’k=-1/2

k=1,-1/2

(x) kxÂ² + kx + 1 = â€“ 4xÂ² â€“ x

Solution:

kxÂ² +4x2+kx+x+1=0

(k+4)x2+(k+1)x+1=0

Here a=k+4, b=k+1, c=1

Discriminant(D)=b2-4ac

=(k+1)2-4*(k+4)*1

=k2+2k+1-4k-16

=k2-2k-15

Roots are real and equal

D=0

k2-2k-15=0

k2-5k+3k-15=0 {Therefore -15=-5*3

-2=-5+3}

k(k-5)+3(k-5)=0

(k-5)(k+3)=0

Either k-5=0, then k=5

or k+3=0, then k=-3

Hence k=5,-3

(xi) (k + 1) xÂ² + 2 (k + 3) x + (k + 8) = 0

Solution:

Here a=k+1, b=2(k+3), c=k+8

Discriminant(D)=b2-4ac

=[2(k+3)]2-4(k+1)(k+8)

=4(k2+6k+9)-4)(k2+9k+8)

=4k2+24k+36-4k2-36k-32

=-12k+4

Roots are real and equal

D=0

-12k+4=0

12k=4â‡’k=4/12=1/3

Hence k=1/3

(xii) xÂ² â€“ 2kx + 7k â€“ 12 = 0

Solution:

Here a=1, b=-2k, c=(7k-12)

Discriminant(D)=b2-4ac

=(-2k)2-4*1*(7k-12)

=4k2-4(7k-12)

=4k2-28k+48

Roots are real and equal

D=0

4k2-28k+48=0

k2-7k+12=0 (Dividing by 4 )

k2-3k-4k+12=0 {12=-3*(-4)

-7=-3-4}

k(k-3)-4(k-3)=0

(k-3)(k-4)=0

Either, k-3=0, then k=3

or k-4=0, then k=4

Therefore, k=3,4

(xiii) (k + 1) xÂ² â€“ 2 (3k + 1) x + 8k + 1 = 0

Solution:

Here a=k+1, b=-2(3k+1), c=8k+1

Discriminant(D)=b2-4ac

=[-2(3k+1)2-4*(k+1)(8k+1)]

=4(9k2+6k+1)-4(8k2+9k+1)

=36k2+24k+4-32k2-36-4

=4k2-12k

Roots are real and equal

D=0

4k2-12k=0

k2-3k=0 ————–(Dividing by 4)

k(k-3)=0

Either k=0

or k-3=0, then k=3

k=3,0

(xiv) 5xÂ² â€“ 4x + 2 + k (4xÂ² â€“ 2x â€“ 1) = 0

Solution:

5x2-4x+2+4kx2-2kx-k=0

(5+4k)x2-(4+2k)x+(2-k)=0

Here a=5+4k, b=-(4+2k), c=2-k

Discriminant (D)=b2-4ac

=[-(4+2k)]2-4*(5+4k)(2-k)

=16+4k2+16k-4(10-5k+8k-4k2)

=16+4k2+16k-40+20k-32k+16k2

=20k2+4k-24

Roots are real and equal

D=0

20k2+4k-24=0

5k2+k-6=0 —–(Dividing by 4)

5k2+6k-5k-6=0

k(5k+6)-1(5k+6)=0

(5k+6)(k-1)=0

Either 5k+6=0, then 5k=-6â‡’k=-6/5

or k-1=0, then k=1

k=1,

(xv) (4 â€“ k) xÂ² + (2k + 4) x (8k + 1) = 0

Solution:

Here a=4-k, b=2k+4, c=8k+1

Discriminant (D)=b2-4ac

=(2k+4)2-4*(4-k)(8k+1)

=4k2+16k+16-4(32k+4-8k2-k)

=4k2+16k+16-128k-16+32k2+4k

=36k2-108k

Roots are real and equal

36k2-108k=0

k2-3k=0â‡’k(k-3)=0

Either k=0

or k-3=0, then k=3

Hence k=0,3

(xvi) (2k + 1) xÂ² + 2 (k + 3) x (k + 5) = 0

Solution:

Here a=2k+1, b=2(k+3), c=k+5

Discriminant (D)=b2-4ac

=[2(k+3)]2-4(2k+1)(k+5)

=4(k2+6k+9)-4(2k2+10k+k+5)

=4k2+24k+36-8k2-40k-4k-20

=-4k2-20k+16

Roots are real and equal D=0

-4k2-20k+16=0

k2+5k-4=0 ———-(Dividing by -4)

Here a=1, b=5, c=-4

Discriminant (D)=b2-4ac

=(5)2-4*1*(-4)=25+16=41

(xvii) 4xÂ² â€“ 2 (k + 1) x + (k + 4) = 0

Solution:

Here a=4, b=-2(k+1), c=k+4

Discriminant (D)=b2-4ac

=[-2(k+1)]2-4*4*(k+4)

=4(k2+2k+1)-16(k+4)

=4k2+8k+4-16k-64

=4k2-8k-12

Roots are real and equal

D=0

4k2-8k-60=0

k2-2k-15=0 ————-(Dividing by 4)

k2-5k+3k-15=0

k(k-5)+3(k-5)=0

(k-5)(k+3)=0

Either k-5=0, then k=5

or k+3=0, then k=-3

k=5,-3

(xviii) 4xÂ² (k + 1) x + (k + 1) = 0

Solution:

Here a=4, b=-2(k+1), c=k+1

Discriminant (D)=b2-4ac

=[-2(k+1)]2-4*4*(k+1)

=4(k2+2k+1)-16(k+1)

=4k2+8k+4-16k-16

=4k2-8k-12

k2-2k-3=0 ————(Dividing by 4)

k2-3k2+k-3=0

k(k-3)+(k-3)=0

(k-3)(k+1)=0

Either (k-3)=0, then k=3

or (k+1)=0, then k=-1

k=-1,3

### Question 3. In the following, determine the set of values of k for which the given quadratic equation has real roots:

(i) 2xÂ² + 3x + k = 0

Solution:

Here a=2, b=3, c=k

Discriminant (D)=b2-4ac

=(3)2-4*2*k

=9-8k

The roots are real

Dâ‰¥0

9-8kâ‰¥0â‡’9â‰¥8kâ‡’8kâ‰¤9

kâ‰¤9/8

(ii) 2xÂ² + x + k = 0

Solution:

Here a=2, b=1, c=k

Discriminant (D)=b2-4ac

=(1)2-4*2*k

=1-8k

The roots are real

Dâ‰¥0

1-8kâ‰¥0â‡’1â‰¥8k

8kâ‰¤1

kâ‰¤1/8

(iii) 2xÂ² â€“ 5x â€“ k = 0

Solution:

Discriminant (D)=b2-4ac

=(-5)2-4*2*(-k)

=25+8k

Roots are real

Dâ‰¥0

25+8kâ‰¥0

8kâ‰¥-25â‡’â‰¥-25/8

kâ‰¥-25/8

(iv) kxÂ² + 6x + 1 = 0

Solution:

Discriminant (D)=b2-4ac

=(6)2-4*k*1

=36-4k

Roots are real

Dâ‰¥0â‡’36-4kâ‰¥0

36â‰¥4kâ‡’4kâ‰¤36

kâ‰¤36/4â‡’kâ‰¤9

kâ‰¤9

(v) 3xÂ² + 2x + k = 0

Solution:

Here a=3, b=2, c=k

Discriminant (D)=b2-4ac

=(2)2-4*3*k=4-12k

Roots are real

Dâ‰¥0â‡’4-12kâ‰¥0

4â‰¥12kâ‡’12kâ‰¤0

4â‰¥12kâ‡’12kâ‰¤4

kâ‰¤â‡’kâ‰¤1/3

### Question 4. Find the values of k for which the following equations have real and equal roots :

(i) xÂ²- 2(k + 1) x + kÂ² = 0

Solution:

Here a=1, b=2(k+1), c=k2

Discriminant (D)=b2-4ac

=[2(k+1)]2-4*1*k2

=4(k2+2k+1)-4k2

=4k2+8k+4-k2

=8k+4

Roots are real and equal

D=0

8k+4=0â‡’8k=-4

k=-4/8=-1/2, Hence k=-1/2

(ii) kÂ²xÂ² â€“ 2 (2k â€“ 1) x + 4 = 0

Solution:

Here, a=k2, b=-2(2k-1), c=4

Discriminant (D)=b2-4ac

=[-2(2k-1)]2-4*k2*4

=4(4k2-4k+1)-16k2

=16k2-16k+4-16k2=-16k+4

Roots are real and equal

D=0

-16k+4=0â‡’-16k=-4

k=4/16=1/4

k=1/4

(iii) (k + 1) xÂ² â€“ 2(k â€“ 1) x + 1 = 0

Solution:

Here, a=k+1, b=-2(k-1) and c=1

Discriminant (D)=b2-4ac

=[-2(k-1)]2-4(k+1)*1

=4(k2-2k+1)-4(k+1)

=4k2-8k+4-4k-4=4k2-12k

Roots are real and equal

D=0

4k2-12k=0

k2-3k=0 ————-(Dividing by 4)

Either k=0

or k-3=0, then k=3

k=0,3

(iv) xÂ² + k(2x + k â€“ 1) + 2 = 0

Solution:

Here, a=1, b=2k, c=(k2-k+2)

Discriminant (D)=b2-4ac

=(2k)2-4*1*(k2-k+2)

=4k2-4k2+4k-8

=4k-8

Roots are real and equal

D=0

4k-8=0â‡’k=2

Hence, k=2

### Question 5. Find the values of k for which the following equations have real roots

(i) 2xÂ² + kx + 3 = 0

Solution:

Here a = 2, b = k, c = 3

Roots are real and equal

D=0

k2-24=0â‡’k2=24

k=Â±âˆš24=Â±âˆš4*6=Â±2âˆš6

(ii) kx (x â€“ 2) + 6 = 0

Solution:

kx2-2kx+6=0

here, a=k, b=-2k, c=6

Discriminant (D)=b2-4ac=(-2k)2-4*k*6=4k2-24k

Roots are real and equal

D=0

4k2-24k=0â‡’4k(k-6)=0

k(k-6)=0

Either k=0 or

k-6=0, then k=6

k=0,6

(iii) xÂ² â€“ 4kx + k = 0

Solution:

Here, comparing with ax2+kx+c=0

a=1, b=-4k, c=k

Discriminant (D)=b2-4ac=(-4k)2-4*1*k=16k2-4k

Roots are real and equal

D=0

16k2-4k=0â‡’4k(4k-1)=0

k(4k-1)=0

Either k=0

or 4k-1=0â‡’4k=1

k=1/4, Hence k=0,1/4

(iv) kx(x â€“ 2âˆš5) + 10 = 0

Solution:

Here a=k, b=-2âˆš5k, c=10

D=b2-4ac

=(-2âˆš5k)2-4(k)(10)=20k2-40k

Roots are equal D=0

20k2-40k=0

k-2=0 ———(Dividing by 20k)

k=2

(v) kx (x â€“ 3) + 9 = 0

Solution:

Here, a=k, b=-3k, c=9

D=b2-4ac

=(-3k)2-4(k)9

=9k2-36k

For roots to be real

D=0

9k2-36k=0

9k(k-4)=0

k-4=0â‡’k=4

k=4

(vi) 4xÂ² + kx + 3 = 0

Solution:

Here, a=4, b=k, c=3

D=b2-4ac

=k2-4(4)(3)

=k2-48

For roots to be real

D=0

k2-48=0

k2=48

k=Â±âˆš48=Â±

k=Â±

### Question 6. Find the values of k for which the given quadratic equation has real and distinct roots :

(i) kxÂ² + 2x + 1 = 0

Solution:

Here, a=k, b=2, c=1

D=b2-4ac

=(2)2-4*k*1

=4-4k

Roots are real and distinct

D>0â‡’4-4k>0

1-k>0â‡’1>k

â‡’k<1

Therefore, k<1

(ii) kxÂ² + 6x + 1 = 0

Solution:

Here, a=k, b=6, c=1

D=b2-4ac

=(6)2-4*k*1

=36-4k

Roots are real and distinct

D>0â‡’36-4k>0

9-k>0â‡’9>k

â‡’k<9

Therefore, k<9

### Question 7. For what value of k, (4 â€“ k) xÂ² + (2k + 4) x + (8k + 1) = 0, is a perfect square.

Solution:

(4 â€“ k) xÂ² + (2k + 4) x + (8k + 1) = 0

Here, a = 4 â€“ k, b = 2k + 4, c = 8k + 1

=(2k+4)2-4*(4-k)(8k+1)

=4k2+16k+16-4(32k+4-8k2-k)

=4k2+16k+16-4(-8k2+31k+4)

=4k2+16k+16+32k2-124k-16

=36k2-108k

Therefore, the given quadratic equation is a perfect square

The roots are real and equal

D=0â‡’36k2-108k=0

Either k=0

or k-3=0â‡’k=3

k=0,3

### Question 8. Find the least positive value of k for which the equation xÂ² + kx + 4 = 0 has real roots.

Solution:

xÂ² + kx + 4 = 0

Here, a=1, b=k, c=4

Therefore, Discriminant(D)=b2-4ac

=(k)2=4*1*4

=k2-16

It has real roots

Dâ‰¥0â‡’k2-16â‰¥0

â‡’k2â‰¥16â‡’(k)2â‰¥(Â±4)2

kâ‰¥4 or kâ‰¤-4

Least positive value of k=4

### Question 9. Find the value of k for which the quadratic equation (3k + 1) xÂ² + 2(k + 1) x + 1 = 0 has equal roots. Also, find the roots.

Solution:

(3k + 1) xÂ² + 2(k + 1) x + 1 = 0

Here a=(3k+1), b=2(k+1) ,c=1

Now, b2-4ac=[2(k+1)]2-4*(3k+1)*1

=4(k2+2k+1)-4(3k+1)

=4k2+8k+4-12k-4

=4k2-4k

Roots are real and equal

b2-4ac=0

4k2-4k=0

k2-k=0 k(k-1)=0

Either k=0 or k-1=0, then k=1

k=0,1

(i) When k=0, then

(3*0*1)x2+2(0+1)x+1=0

x2+2x+1=0

(x+1)2=0

x+1=0

x=-1

When k=1, then

(3*1+1)x2+2(1+1)x+1=0

4x2+4x2+1=0

(2x+1)2=0

2x+1=0

2x=-1â‡’x=-1/2

x=1, -1/2

### Question 10. Find the values of p for which the quadratic equation (2p + 1) xÂ² â€“ (7p + 2) x + (7p â€“ 3) = 0 has equal roots. Also, find these roots.

Solution:

Here, a=2-+1, b=-(7p+2), c=(7p-3)

D=0 [Equal roots]

As b2-4ac=0

[-(7p+)]2-4(2p+1)(7p-3)=0

(7p+2)2-4(14p2-6p+7p-3)=0

49p2+28p+4-56p2+24p-28p+12=0

-7p2+24p+16=0

7p2-24-16=0 ————-(Dividing both sides by -1)

7p(p-4)+4(p-4)=0

(p-4)(7p+4)=0

p-4=0 or 7p+4=0

p=4 or p=-4/7

Roots are x=-b/2a [As equal roots (given)]

Where p=4,

Equal roots is 5/3

When p=-4/7

Equal roots are 5/3 and 7

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