 # RS Aggarwal Class 10 Solutions Chapter 10 - Quadratic Equations Ex 10F ( 10.6)

## RS Aggarwal Class 10 Chapter 10 - Quadratic Equations Ex 10F ( 10.6) Solutions Free PDF

Scoring better marks in the Class 10 examination is important for CBSE board students. The RS Aggarwal Class 10 solutions are beneficial for students in fetching more marks in maths subject. The solutions are created according to the latest syllabus of the CBSE. Students should refer to these solutions provided by us to turn challenging topics into an easy one. All the solutions are arranged systematically in a proper step by step format.

Practicing from these solutions will help students to secure higher marks. It will help you in clearing all the doubts of the topics as the solutions cover the complete syllabus. The solutions teach students to solve the questions in a simple and easy way. The RS Aggarwal Solutions Class 10 Chapter 10 – Quadratic Equations Ex 10.6 are beneficial for students to revise the whole syllabus.

1. Find the nature of the roots of the following quadratic equations:

(i)2x2 – 8x + 5 = 0 (ii) 3x2– 2$\sqrt{6}$x +2 = 0

(iii)5x2-4x+1=0 (iv) 5x(x – 2) + 6 =0

(v) 12x2 – 4$\sqrt{15}$x+5=0 (vi) x2 – x + 2 = 0

(i)The given equation is 2x2 – 8x + 5 = 0.

This is of the form ax2 + bx + c = 0, where a = 2, b = -8 and c = 5.

Therefore, Discriminant, D = b2 – 4ac = (-8)2 – 4 x 2 x 5 = 64 – 40 = 24> 0

Hence, the given equation has real and unequal roots.

(ii) The given equation is 3x2 — 2$\sqrt{6}$x + 2 = 0.

This is of the form ax2 + bx + c = 0, where a = 3, b = -2$\sqrt{6}$ and c = 2.

Therefore, Discriminant, D= b2 – 4ac = (-2$\sqrt{6}$)2 -4 x 3 x 2 = 24 – 24 = 0

Hence, the given equation has real and equal roots.

(iii) The given equation is 5x2 – 4x + 1 = 0.

This is of the form ax2 + bx + c = 0, where a = 5, b = -4 and c = 1.

Therefore, Discriminant, D = b2 – 4ac = (-4)2 – 4 x 5 x 1 = 16 – 20 = -4 < 0

Hence, the given equation has no real roots.

(iv) The given equation is

5x(x – 2) + 6 = 0

$\Rightarrow$ 5x2 – 10x + 6 = 0

This is of the form ax2 + bx + c = 0, where a = 5, b = -10 and c = 6.

Therefore, Discriminant, D = b2 – 4ac = (-10)2 – 4 x 5x 6 = 100 -120 = -20 < 0

Hence, the given equation has no real roots.

(v) The given equation is 12x2 – 4$\sqrt{15}$x + 5 = 0.

This is of the form ax2 + bx + c = 0, where a = 12, b = -4$\sqrt{15}$ and c = 5.

Therefore, Discriminant, D = b2 – 4ac = (-4$\sqrt{15}$)2 – 4 x 12 x 5 = 240 – 240 = 0

Hence, the given equation has real and equal roots.

(vi) The given equation is x2 – x + 2 = 0

This is of the form ax2 + bx + c = 0, where a=1, b=-1 and c=2.

Therefore, Discriminant, D = b2 – 4ac = (-1)2 – 4 x 1 x2 = 1 – 8 = -7<0

Hence, the given equation has no real roots.

2. If a and b are distinct real numbers, show that the quadratic equation

2(a2+ b2)x2 + 2(a+ b)x + 1 =0 has no real roots.

The given equation is 2 (a2 + b2) x2 + 2(a + b)x + 1= 0.

Therefore, D = [2(a+ b)]2 – 4 x 2(a2 + b2) x 1

= 4(a2 + 2ab+ b2) – 8 (a2 + b2)

= 4a2 + 8ab+ 4b2 – 8a2 – 8b2

= -4a2 + 8ab – 4b2

= -4(a2 — 2ab+ b2)

= -4(a — b)2<0

Hence. the given equation has no real roots.

3. Show that the roots of the equation x2 + px-q2= 0 are real for all real values of p and q.

Given:

x2 ± px – q2 =0

Here,

a= 1, b = p and c = -q2

Discriminant D is given by :

D = (b2 – 4ac)

= p2 – 4 x 1 x (-q2)

= (p2+4q2) > 0

D > 0 for all real values of p and q .

Thus, the roots of the equation are real.

4. For what values of k are the roots of the quadratic equation 3x2+ 2kx + 27 = 0 real and equal?

Given:

3x2 + 2kx + 27 = 0

Here,

a = 3, b = 2k and c = 27

It is given that the roots of the equation are real and equal; therefore, we have:

D = 0

$\Rightarrow$ (2k)2 – 4 x 3 x 27 = 0

$\Rightarrow$4k2 – 324 = 0

$\Rightarrow$4k2 = 324

$\Rightarrow$k2 = 81

$\Rightarrow$k = ±9

Therefore, k = 9 or k = -9

5. For what value of k are the roots of the quadratic equation ix(x-2$\sqrt{5}$) + 10 = 0 real and equal?

The given equation is

Kx(x- 2$\sqrt{5}$) + 10 = 0

$\Rightarrow$kx2 — 2$\sqrt{5}$kx + 10 = 0

This is of the form ax2 + bx + c=0, c =0, where a = k, b = -2$\sqrt{5}$k and c= 10.

Therefore, = b2 – 4ac=(-2$\sqrt{5}$k)2 – 40K

The given equation will have reaI and equal roots if D = 0.

Therefore, 20k2 – 40k = 0

$\Rightarrow$ 20k(k – 2) = 0

$\Rightarrow$ k = 0 or k – 2 = 0

$\Rightarrow$k = 0 or k = 2

But, for k = 0, we get 10 = 0, which is not true.

Hence, 2 is the required value of k.

6. For what values of p are the roots of the equation 4x2 + px + 3 = 0 real and equal?

The given equation is 4x2 px + 3 = 0.

This is of the form ax2 + bx + c = 0, where a = 4, b = p and c = 3.

Therefore, D = b2 — 4ac = p2 – 4 x 4 x 3 =p2 – 48

The given equation will have real and equal roots if D = 0.

Therefore, p2 – 48 = 0

$\Rightarrow$p2 = 48

$\Rightarrow$p= ±4$\sqrt{3}$

Hence, 4$\sqrt{3}$ and -4$\sqrt{3}$ are the required values of p.

7. Find the nonzero value of k for which the roots of the quadratic equation

9x2-3kx + k = 0 are real and equal.

The given equation is 9x2 – 3kx + k = 0.

This is of the form ax2 + bx + c = 0, where a = 9, b = -3k and c = k.

Therefore, D = b2 – 4ac (-3k)2 – 4 x 9 x k = 9k2 – 36k

The given equation will have real and equal roots if D = 0.

Therefore, 9k2 – 36k = 0

$\Rightarrow$9k(k – 4) = 0

$\Rightarrow$k = 0 or k – 4 = 0

$\Rightarrow$k = 0 or k = 4

But, k $\neq$ 0 (Given)

Hence, the required value of k is 4.

8. Find the values of k for which the quadratic equation (3k+ 1)x2 + 2(k + 1)x + 1 = 0 has real and equal roots.

The given equation is (3k + 1)x2 + 2(k + 1)x + 1= 0.

This is of the form ax2 + bx+ c= 0, where a = 3k +1, b = 2(k + 1) and = 1.

Therefore, D = b2 – 4ac

= [2(k +1)]2 — 4 x (3k + 1) x 1

= 4(k2 + 2k + 1) — 4(3k + 1)

= 4k2 + 8k + 4 – 12k – 4

= 4k2 — 4k

The given equation will have real and equal roots if D = 0.

Therefore, 4k2 – 4k = 0

$\Rightarrow$4k(k -1) = 0

$\Rightarrow$k = 0 or k —1= 0

$\Rightarrow$k= 0 or k =1

Hence, 0 and 1 are the required values of k.

9. Find the values of p for which the quadratic equation

(2P+1)x2 -(7p + 2) x + (7p – 3) = 0 has real and equal roots.

The given equation is (2p+ 1)x2 — (7p+ 2)x + (7p — 3) = 0.

This is of the form ax2 + bx + c = 0, where a = 2p +1, b = -(7p + 2) and c = 7p- 3.

Therefore, D = b2 — 4ac

= [—(7p + 2)]2 — 4 x (2p + 1) x (7p — 3)

= (49p2 + 28p + 4) — 4(14P2 + p – 3)

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

= -7p2 + 24p+ 16

The given equation will have real and equal roots if D= 0.

Therefore, -7p2 + 24p+ 16 = 0

$\Rightarrow$7p2 — 24p — 16 = 0

$\Rightarrow$7p2 – 28p + 4p — 16 = 0

$\Rightarrow$7p(p — 4) + 4(p — 4) = 0

$\Rightarrow$ (p — 4)(7p+ 4) = 0

$\Rightarrow$p-4 =0 or 7p+4=0

$\Rightarrow$p=4 p=$-\frac{4}{7}$

Hence, 4 and $-\frac{4}{7}$ are the required values of p.

10. Find the values of p for which the quadratic equation

(9+1)x2 – 6(p + 1) x + 3(p + 9) = 0, p $\neq$ -1 has equal roots. Hence, find the root of

the equation.

The given equation is (p + 1)x2 — 6(p + 1)x + 3(p + 9) = 0.

This is of the form ax2 + bx + c = 0, where a = p +1, b = -6(p + 1) and c = 3(p + 9).

Therefore, D = b2 – 4ac

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

= 12(p + 1)13(p + 1) — (p + 9)]

= 12(p + 1)(2p — 6)

The given equation will have real and equal roots if D = 0.

Therefore, 12(p + 1)(2p – 6)=0

$\Rightarrow$p+1 =O or 2p-6=0

$\Rightarrow$p=-1 o p=3

But, p $\neq$ -1 (Given)

Thus, the value of p is 3.

Putting p = 3, the given equation becomes 4x2 — 24x + 36 = 0.

$\Rightarrow$4x2 — 24x + 36 = 0

$\Rightarrow$4(x2 — 6x + 9) = 0

$\Rightarrow$(x — 3)2 = 0

$\Rightarrow$x — 3 = 0

$\Rightarrow$x = 3

Hence, 3 is the repeated root of this equation.

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

It is given that -5 is a root of the quadratic equation 2x2+ px —15 = 0.

Therefore, 2(-5)2 -p x (-5) — 15 = 0

$\Rightarrow$-5p+ 35 = 0

$\Rightarrow$p = 7

The roots of the equation px2 – px + k = 0 = 0 are equal.

Therefore, D=0

$\Rightarrow$p2 – 4pk = 0

$\Rightarrow$ (7)2-4 x 7 x k=0

$\Rightarrow$49 – 28k = 0

$\Rightarrow$k=$\frac{49}{28}$= $\frac{7}{4}$

Thus, the value of k is $\frac{7}{4}$

12. If 3 Is a root of the quadratic equation x2 – x + k = 0, find the value of P so that the root of the equation x2 + k (2x + k + 2)+p=0 are equal.

It is given that 3 is a root of the quadratic equation x2 – x + k = 0.

Therefore, (3)2 -3+ k = 0

$\Rightarrow$k+6=0

$\Rightarrow$k = -6

The roots of the equation x2 + 2kx + (k2 + 2k +p) = 0 are equal.

Therefore, D=0

$\Rightarrow$ (2k)2 – 4 x 1 x (k2 + 2k +p) = 0

$\Rightarrow$ 4k2 – 4k2 – 8k- 4p = 0

$\Rightarrow$ -8k- 4p= 0

$\Rightarrow$p= $\frac{8k}{-4}$ =-2k

$\Rightarrow$p=-2x(-6)=12

Hence, the value of p is 12.

13. If -4 is a root of the equation x2 + 2x + 4p = 0, find the value of k for which the quadratic equation x2 + px +(1+ 3k) + 7(3 + 2k)= 0 has equals roots.

It is given that —4 is a root of the quadratic equation x2 + 2x + 4p = 0.

Therefore, (-4)2 +2 x (-4)+4p=0

$\Rightarrow$16— 8+4p= 0

$\Rightarrow$4p+8= 0

$\Rightarrow$p=-2

The equation x2+px(1 + 3k) + 7(3 + 2k) = 0 has equal roots.

Therefore, D = 0

$\Rightarrow$ [p(1 + 3k)]2 — 4 x 1 x 7(3 + 2k) = 0

$\Rightarrow$ [-2(1 + 3k)]2 — 28(3 + 2k) = 0

$\Rightarrow$4(1 + 6k + 9k2) — 28(3 + 2k) = 0

$\Rightarrow$4(1 + 6k + 9k2 — 21 — 14k) =0

$\Rightarrow$9k2 — 8k — 20 = 0

$\Rightarrow$9k2 — 18k + 10k — 20 = 0

$\Rightarrow$9k(k — 2) + 10(k — 2) = 0

$\Rightarrow$ (k — 2)(9k + 10) = 0

$\Rightarrow$k — 2=0 or 9k+ 10 =0

$\Rightarrow$k = 2 or k = —$-\frac{10}{9}$

Hence. the required value of k is 2 or —$-\frac{10}{9}$

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

Given:

(1 + m2)x2 + 2mcx + (c2 – a2) = 0

Here,

a= (1 + m2), b = 2mc and c = (c2 — a2)

It is given that the roots of the equation are equal; therefore, we have:

D = 0

$\Rightarrow$ (b2 — 4ac) = 0

$\Rightarrow$ (2mc)2 — 4 x (1 + m2) x (c2 — a2) = 0

$\Rightarrow$ 4m2c2 – 4(c2 – a2 + m2c2 – m2a2) = 0

$\Rightarrow$4m2c2 – 4c2 + 4a2 + 4m2c2 + 4m2a2 = 0 =

$\Rightarrow$ -4c2 + 4a2 + 4m2a2 = 0

$\Rightarrow$ a2 + m2a2 = c2

$\Rightarrow$a2 (1 + m2) = c2

$\Rightarrow$c2 = a2(1 + m2)

Hence proved.

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

Given:

(c2 — ab)x2 — 2(a2 – bc)x + (b2 – ac) = 0

Here,

a = (c2 – ab), b = — 2(a2 — bc), c = (b2 — ac)

It is given that the roots of the equation are real and equal; therefore, we have:

D=0

$\Rightarrow$ (b2 — 4ac) = 0

$\Rightarrow${-2(a2 – bc)}2 – 4 x (c2 – ab) x (b2 — ac) = 0

$\Rightarrow$4(a4 — 2a2bc + b2c2) – 4(b2c2 – ac3 — ab3 + a2bc) = 0

$\Rightarrow$a4 — 2a2bc – b2c2 — b2c2 + ac3 + ab3 — a2bc = 0

$\Rightarrow$a4 — 3a2bc + ac3 + ab3 = 0

$\Rightarrow$a(a3 — 3abc + c3 + b3) = 0

Now,

a = 0 or a3 — 3abc + c3 + b3 = 0

a = 0 or a3 + b3 + c3 = 3abc

16. Find the values of p for which the quadratic equation 2x2 + px +8=0 has real roots.

Given:

2x2 + px + 8 = 0

Here,

a = 2, b = p and c = 8

Discriminant D is given by :

D = (b2 – 4ac)

= p2 — 4 x 2 x 8

= (p2 — 64)

If D $\geq$ 0, the roots of the equation will be real.

$\Rightarrow$ (p2 — 64) $\geq$ 0

$\Rightarrow$ (p + 8) (p — 8) $\geq$ 0

$\Rightarrow$p $\geq$8 and p $\leq$-8

Thus, the roots of the equation are real for p $\geq$ 8 and p $\leq$ -8.

17. Find the value of a for which the equation (a – 12)x2 + 2 (a – 12)x+2=0 has equal roots.

Given:

(a — 12)x2 + 2(a — 12)x + 2 = 0

Here,

a = (a — 12), b = 2($\alpha$ — 12) and c = 2

It is given that the roots of the equation are equal; therefore, we have:

D = 0

$\Rightarrow$ (b2 — 4ac) = 0

$\Rightarrow$ {2($\alpha$ — 12)}2 — 4 x ($\alpha$ — 12) x 2 = 0

$\Rightarrow$ 4{$\alpha$2 — 24$\alpha$ + 144) — 8($\alpha$ — 12) = 0

$\Rightarrow$4$\alpha ^{2}$ — 96$\alpha$ + 576 — 8$\alpha$ + 96 = 0

$\Rightarrow$4$\alpha ^{2}$ — 104$\alpha$ + 672 = 0

$\Rightarrow$ $\alpha ^{2}$ — 26$\alpha$ + 168 = 0

$\Rightarrow$ $\alpha ^{2}$ — 14$\alpha$ — 12$\alpha$ + 168 = 0

$\Rightarrow$ $\alpha$ ($\alpha$ — 14) — 12($\alpha$ — 14) =0

$\Rightarrow$ ($\alpha$ — 14)( $\alpha$ — 12) =0

Therefore, $\alpha$ = 14 or $\alpha$ = 12

If the value of a is 12, the given equation becomes non-quadratic. Therefore, the value of $\alpha$ will be 14 for the equation to have equal roots.

18. Find the value of k for which the roots of 9x2 + 8kx + 16 = 0 are real equal.

Given:

9x2 + 8kx + 16 = 0

Here,

a = 9, b = 8k and c = 16

It is given that the roots of the equation are real and equal; therefore, we have:

D = 0

$\Rightarrow$ (b2 — 4ac) = 0

$\Rightarrow$ (8k)2 — 4 x 9 x 16 = 0

$\Rightarrow$64k2 — 576 = 0

$\Rightarrow$64k2 = 576

$\Rightarrow$k2 = 9

$\Rightarrow$k= ±3

Therefore, k = 3 or k = —3

19. Find the values of k for which the given quadratic equation has real distinct roots:

(i) kx2+6x + 1 = 0 (ii) x2 – kx + 9 = 0

(iii)9x2+3kx + 4 = 0 (iv) 5x2– kx + 1 =0

(i) The given equation is kx2 -I- 6x +1= 0

Therefore, D=62-4xkx1=36-4k

The given equation has real and distinct roots if D > 0.

Therefore, 36-4k>0

$\Rightarrow$4k < 36

$\Rightarrow$k<9

(ii) The given equation is x2 — kx + 9 = 0.

Therefore, D=(—k)2-4x1x9=k2-36

The given equation has real and distinct roots if D > 0.

Therefore, k2-36>0

$\Rightarrow$ (k — 6)(k + 6) > 0

$\Rightarrow$k<-6 or k>6

(iii) The given equation is 9x2 +3kx-1+4 = 0.

Therefore, D = (3k)2 — 4 x 9 x 4 = 9k2 — 144

The given equation has real and distinct roots if D> 0.

Therefore, 9k2 — 144 > 0

$\Rightarrow$9 (k2 — 16) > 0

$\Rightarrow$ (k — 4)(k + 4) > 0

$\Rightarrow$k < —4 or k > 4

(iv) The given equation is 5x2 — kx+ 1= 0.

Therefore, D=(—k)2-4x 5 x 1=k2 —20

The given equation has real and distinct roots if D> 0.

Therefore, k2-20 >0

$\Rightarrow$k2 – (2$\sqrt{5}$)2> 0

$\Rightarrow$ (k-2$\sqrt{5}$)(k+2$\sqrt{5}$) >0

$\Rightarrow$k < —2$\sqrt{5}$ or k > 2$\sqrt{5}$

20. If a and b are real and a $\neq$ b then show that the roots of the equation (a-b)x2+5(a +b)x – 2(a – b) = 0 are real and unequal.

The given equation is

(a — b)x2 + 5(a + b)x — 2(a — b) = 0.

Therefore, D = [5(a + b)]2 — 4 x (a — b) x [ —2(a — b)]

= 25(a + b)2 + 8(a — b)2

Since a and b are real and a $\neq$ b, so (a — b)2> 0 and (a +b)2> 0.

Therefore, 8(a —b)2> 0 ……….. (1) (Product of two positive numbers is always positive)

Also, 25(a + b)2> 0 …………(2) (Product of two positive numbers is always positive)

Adding (1) and (2), we get

25(a + b)2 + 8(a — b)2 > 0 (Sum of two positive numbers is always positive)

$\Rightarrow$D > 0

Hence, the roots of the given equation are real and unequal.

21. If the roots of the equation (a2+b2)x2 – 2(ac + bd)x+ (c2+d2)=0 are equal, prove that $\frac{a}{b}$= $\frac{c}{d}$.

It is given that the roots of the equation (a2 + b2)x2 — 2(ac + bd)x + (c2 + d2) = 0 are equal.

Therefore, D=0

$\Rightarrow$ [-2(ac+ bd)]2 — 4 (a2 + b2) (c2+d2) =0

$\Rightarrow$ 4(a2c2 +b2d2 + 2abcd) — 4(a2 c2 + a2d2 + b2c2 + b2d2) = 0

$\Rightarrow$ 4 (a2 c2 + b2 d2 + 2abcd — a2e — a2d2 — b2 c2 – b2d2) =0

$\Rightarrow$ (—a2d2 + 2abcd — b2c2) = 0

$\Rightarrow$— (a2d2 — 2abcd + b2 c2) = 0

$\Rightarrow$ (ad – bc)2 = 0

$\Rightarrow$ad— bc = 0

$\Rightarrow$ad= bc

$\Rightarrow$ $\frac{a}{b}$= $\frac{c}{d}$

Hence Proved.

22. If the roots of the equations ax2+2bx+c=0 and bx2-2$\sqrt{ac}$x + b = 0 are simultaneously real then prove that b2=ac.

It is given that the roots of the equation ax2 + 2bx + c = 0 are real.

Therefore, D1=(26)2-4xaxc$\geq$0

$\Rightarrow$4(b2 — ac) $\geq$ 0

$\Rightarrow$b2 — ac $\geq$ 0 …………. (1)

Also, the roots of the equation bx2 — 2 $\sqrt{acx}$ +b = 0 are real.

Therefore, D2=(-2$\sqrt{ac}$2-4xbxb $\geq$0

$\Rightarrow$4(ac — b2) $\geq$0

$\Rightarrow$—4 (b2 — ac) $\geq$ 0

$\Rightarrow$b2—ac$\leq$0 ………………..(2)

The roots of the given equations are simultaneously real if (1) and (2) holds true together. This is possible if

b2 — ac = 0

$\Rightarrow$bb2= ac

### Key Features of RS Aggarwal Class 10 Solutions Chapter 10 – Quadratic Equations Ex 10F ( 10.6)

• It provides easy solutions to difficult and tricky questions.
• It is explained in a step by step manner for better understanding.
• It helps in clearing your doubts and concepts.
• Solving RS Aggarwal Class 10 solutions help students in boosting their confidence level.

#### Practise This Question

A real image of half the size is obtained in a concave spherical mirror with a radius of curvature of 40 cm. The distance of the object and that of its image from the mirror will be respectively