Skip to content
Related Articles

Related Articles

Improve Article

Class 10 RD Sharma Solutions – Chapter 8 Quadratic Equations – Exercise 8.6 | Set 2

  • Last Updated : 16 May, 2021

Question 11. If – 5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal-roots, find the value of k. 

Solution:

2x² + px – 15 = 0

-5 is its one root

It will satisfy it

2(-5)2+p(-5)-15=0



2*25-5p-15=0

50-15-5p=0

-5p+35=0

-5p=-35

p=-35/-5=7

Now in equation 

p(x2+x)+k=0

7(x2+x)+k=0

7x2+7x+k=0

Here, a=7, b=7, c=k

D or b2-4ac = (7)2-4*7*k

=49-28k

Roots are real and equal

D or b2-4ac=0

49-28k=0⇒28k=49

k=49/28=7/4

Question 12. If 2 is a root of the quadratic equation 3x² + px – 8 = 0 and the quadratic equation 4x² – 2px + k = 0 has equal roots, find the value of k. 

Solution:

2 is a root of 3x2+px-8=0



It will satisfy if

3(2)2+p*2-8=0

12+2p-8=0

4+2p=0

2p=-4

p=-4/2=-2

p=-2

Now in equation,

4x2-2px+k=0

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

4x2+4x+k=0

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

D=b2-4ac=(4)2-4*4*k

=16- 16k

Roots are real and equal

D or b2-4ac=0

 16-16k=0

=> 16k = 16

k = 16

Question 13. If 1 is a root of the quadratic equation 3x² + ax – 2 = 0 and the quadratic equation a(x² + 6x) – b=0 has equal roots, find the value of b.

Solution:

1 is one root of

3x² + ax – 2 = 0

3(1)2+a*1-2=0

3+a-2=0⇒a+1=0

a=-1

Now in equation a(x2+6x)-b=0

-1(x2+6x)-b=0

-x2-6x-b=0

⇒x2+6x+b=0

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



D=b2-4ac=(6)2-4*1*k

=36-4k

Roots are equal

D or b2-4ac=0

36-4k=0

4k=-36

k=-36/4=-9

Question 14. Find the value of p for which the quadratic equation (p + 1) x² – 6 (p + 1) x + 3 (p + q) = 0, p ≠ -1 has equal roots. Hence, find the roots of the equation. 

Solution:

(p+1)x2-6(p+1)x+3(p+9)=0,

p≠-1

Comparing with ax2+bx+c=0

b2-4ac, c=p+1, b=-6(p+1), c=3(p+9)

Now D=b2-4ac

=[-6(p+1)]2-4*(p+1)*3(p+9)

=36(p+9)2-12(p+1)(p+9)

Roots are equal

D=0

⇒36(p+1)2-12(p+1)(p+9)=0

⇒36(p+1)2=12(p+1)(p+9)

⇒3(p+1)=p+9⇒3p+3=p+9

⇒3p-p=⇒9-3⇒2p=6

p=6/2=3

p=3

x=\frac{-b±\sqrt{b^2-4ac}}{2a}\\ =\frac{6(p+1)+0}{2*(p+1)}=\frac{6(p+1)}{2(p+1)}=3

Hence, x=3,3

Question 15. Determine the nature of the roots of the following quadratic equations :

(i) (x – 2a) (x – 2b) = 4ab

Solution:

 ⇒x2-2bx-2ax+4ab-4ab=0

Here,  a=1,b=-2(a+b),  c=0

Discriminant(D)=b2-4ac



={-2(a+b)}2-4*1*0

={-2(a+b)}2

D>0

Roots are real and distinct

(ii) 9a²b²x² – 24abcdx + 16c²d² = 0, a ≠ 0, b ≠ 0

Solution:

Here a=9a2b2, b=-24abcd, c=16c2d2

D=b2-4ac

=(-24abcd)2-4*9a2b2*16c2d2

=576a2b2c2d2-576a2b2c2d2

=0

D=0

Roots are real and equal

(iii) 2 (a² + b²) x² + 2 (a + b) x + 1 = 0

Solution:

Here a=2(a2+b2), b=2(a+b), c=1

D=b2-4ac

={2(a+b)}2-4*2(a2+b2)*1

=4(a2+b2+2ab)-8(a2+b2)

=4a2+4b2+8ab-8a2-8b2

=8ab-4a2-4b2

=-(4a2+4b2-8ab)

=-4(a2+b2-2ab)

=-4(a-b)2

D<0

Root are not real

(iv) (b + c) x² – (a + b + c) x + a = 0

Solution:

Here a=b+c, b=-(a+b+c),  c=a

D=b2-4ac

=[-(a+b+c)]2-4*(b+c)*a

=a2+b2+c2+2ab+2bc+2ca=4ab-4ca

=a2+b2+c2-2ab+2bc-2ca

=[-a+b+c]2

D>0

Roots are real and distinct.

Question 16. Determine the set of values of k for which the given following quadratic equation has real roots :

(i) x² – kx + 9 = 0

Solution:

Here a=1, b=-k, c=9

D=b2-4ac



=(-k)2-4*1*9

=k2-36

Roots are real

D≥⇒k2-36≥0

k2≥36⇒k2(±6)2

k≥6  or k≤-6

(ii) 2x² + kx + 2 = 0

Solution:

Here,  a=2, b=k,  c=2

D=b2-4ac

=(k)2-4*2*2

=k2-16

Roots are real

D≥0⇒k2-16≥0

k2≥16⇒(k)2≥(±4)2

k≥4  or  k≤-4

(iii) 4x² – 3kx +1=0

Solution:

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

D=b2-4ac

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

=9k2-16

Roots are real 

D≥0⇒9k2-16≥0

9k2≥16⇒k2≥16/9

(k)2≥(±\frac{4}{3})2

k≥4/3 or k≤-4/3

(iv) 2x² + kx – 4 = 0

Solution:

Here a=2, b=k, c=-4

D=b2-4ac

=k2-4*2*(-4)

=k2+32

The roots are real

D≥0⇒k2+32≥0

k2+32≥0 for all value of 

k ∈ R

Question 17. If the roots of the equation (b – c) x² + (c – a) x + (a – b) = 0 are equal, then prove that 2b = a + c. [CBSE 2002C]

Solution:

(b – c) x² + (c – a) x + (a – b) = 0

Here a=b-c,  B=c-a,  c=a-b

D=b2-4ac

=(c-a)2-4(b-c)(a-b)

The roots are equal

D=0

(c-a)2-4(b-c)(a-b)=0

c2+a2-2ca-4(ab-b2-ca+bc)=0

c2+a2-2ca-4ab+4b2+4ca-4bc=0

a2+4b2+c2-4ab-4bc+2ca=0

(a-2b+c)2=0⇒a-2b+c=0

=> a + c = 2b



=> 2b = a + c

Hence proved.

Question 18. If the roots of the equation (a² + b²) x² – 2 (ac + bd) x + (c² + d²) = 0 are equal. prove that ab = cd

Solution:

Here a=a² + b², b= – 2 (ac + bd), c=c² + d²

D=b2-4ac

=[-2(ac+bd)]2-4(a2+b2)(c2+d2)

=4(ac+bd)2-4[a2c2+a2d2+b2c2+b2d2]

=4[a2c2+b2d2+2abcd]-4[a2c2+a2d2+b2c2+b2d2]

=4a2c2+4b2d2+8abcd-4a2d2-4b2c2-4b2d2

=8abcd-4a2d2-4b2c2

The roots are equal

D=0

8abcd-4a2d2-4b2c2=0

a2d2+b2c2-2abcd=0                 ———–(Dividing by -4)

(ad-bc)2=0⇒ad-bc=0

ad=bc⇒a/b=c/d

Question 19. If the roots of the equations ax² + 2bx + c = 0 and bx² – 2√ac x + b = 0 are simultaneously real, then prove that b² = ac

Solution: 

ax2+2bx+c=0             ———–(i)

and bx2-2\sqrt{ac} x+b=0     ———–(ii)

Let D1 and D2 are the discriminants of there simultaneous equation (i) and (ii)

D1=(2b)2-4*a*c=4b2-4ac

 and D2=(-2\sqrt{ac} )2-4*b*b

=4ac=4b2

These have real roots

D1≥0⇒4b2-4ac≥0

⇒4b2≥4ac⇒b2≥ac         ————-(i)

and D2≥0

Therefore, 4ac-4b2≥0 ⇒4ac≥4b2

ac ≥ b2    ——————(ii)

ac≥b2≥ac



b2=ac

Question 20. If p, q are real and p ≠ q, then show that the roots of the equation (p – q) x² + 5(p + q) x – 2(p – q) = 0 are real and unequal.

Solution:

Here a=p-q, b=5(p+q), c=-2(p-q)

D=b2-4ac

=[5(p+q)]2-4*(p-q)*-2(p-q)

=25(p+q)2+8(p-q)2

p and q are real and p≠q

25(p+q)2+8(p-q)≥0

The given quadratic equation has real and unequal roots.

Question 21. If the roots of the equation (c² – ab) x² – 2 (a² – bc) x + b² – ac = 0 are equal, prove that either a = 0 or a3 + b3 + c3 = 3abc.

Solution:

Here a=c2-ab, b=-2(a2-bc), c=b2-ac

D=b2-4ac

=[-2(a2-bc)]2-4(c2-ab)(b2-ac)

=4[a4+b2c2-2a2bc]-4[b2c2-ac3-ab3+a2bc]

=4a4+4b2c2-8a2bc-4b2c2+4ac3+4ab3-4ac2bc

=4a4+4ab+4ac3-12a2bc

=4a[a3+b3+c3-3abc]

D=0

4a(a3+b3+c3-3abc)=0

a(a3+b3+c3-3abc)=0

Either a=0

or a3+b3+c3=3abc=0

a3+b3+c3=3abc

Question 22. Show that the equation 2 (a² + b²) x² + 2 (a + b) x + 1 = 0 has no real roots, when a ≠ b.

Solution:

Here a=2(a+b2), b=2(a+b), c=1

D=b2-4ac

[2(a+b)]2-4*2*(a2+b2)*1

4(a+b2+2ab)-8(a2+b2)

=4a2+4b2+8ab-8a2-8b2

=-4a2-b2+8ab

-4[a2+b2-2ab]

=-4(a-b)2

D<0

Roots are not real

Question 23. Prove that both the roots of the equation (x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0 are real but they are equal only when a = b = c.

Solution:

Here a=3, b=-2(a+b+c), c=ab+ac+ca

D=b2-4ac

=[-2(a+b+c)]2-4*3(ab+bc+ca)

=4(a+b+c)2-12(ab+bc+ca)

=4[(a+b+c)]2-3(ab+bc+ca)

=4[a2+b2+c2+2ab+2bc+2ca-3ab-3bc-3ca]

=2[(a-b)]2+(b-c)2+(c-a)2]

Clearly, D≥0

Roots are real 

If roots are equal, then

D=0

(a-b)2+(b-c)2+(c-a)2=0

a-b=0, b-c=0, c-a=0

a=b=, b=c, c=a

a=b=c   

Hence proved

Question 24. If a, b, c are real numbers such that ac ≠ 0, then show that at least one of the equations ax² + bx + c = 0 and – ax² + bx + c = 0 has real roots.

Solution:

a,b,c are real number

and ac≠0

ax2+bx+c=0        ———-(i)

-ax+bx+c=0        ——–(ii)

Let D1 and D2 be the discriminants of the two equation (i) and (ii)

D1= b2-4ac and D2=b2-4ac

If equation (i) has real roots, then

D1≥0



b2-4ac≥0

b2-≥ac

Now D2=b2+4ac

4ac≤b2

b2+4ac≥0

D≥0

Both the equation has real roots

Hence proved

Question 25. If the equation (1 + m²) x² + 2mcx + (c² – a²) = 0 has equal roots, prove that c² = a² (1 + m²). 

Solution:  

Here a=1+m2, b=2mc, c=c2-a2

D=b2-4ac

=(2mc)2-4(1+m2)(c2-a2)

=4m2c2-4c2+4a-4m2c2+4m2a2

=4a2+4m2a2-4c2

Root are equal

D=0⇒4a2+4m2a2-4c2=0

a2+m2a2-c2=0

a2+m2a2=c2

a2(a+m2)=c2

c2=a2(1+m2)

Hence proved

Attention reader! Don’t stop learning now. Join the First-Step-to-DSA Course for Class 9 to 12 students , specifically designed to introduce data structures and algorithms to the class 9 to 12 students




My Personal Notes arrow_drop_up
Recommended Articles
Page :