RS Aggarwal Solutions Class 10 Ex 10F

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

Answer :

(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.

Answer :

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.

Answer :

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?

Answer :

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?

Answer :

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?

Answer :

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.

Answer :

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.

Answer :

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.

Answer :

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.

Answer :

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.

Answer :

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.

Answer :

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.

Answer :

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).

Answer :

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 .

Answer :

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.

Answer :

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.

Answer :

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.

Answer :

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

Answer :

(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.

Answer :

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 \(\frac{a}{b}\)= \(\frac{c}{d}\).

Answer :

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.

Answer :

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


Practise This Question

What are the advantages of using a biomass fuel?