Interactive OP Malhotra Maths Class 11 Solutions Chapter 10 Quadratic Equations Ex 10(d) engage students in active learning and exploration.

## S Chand Class 11 ICSE Maths Solutions Chapter 10 Quadratic Equations Ex 10(d)

Question 1.

Find the condition that one root of ax² + bx + c = 0 may be

(i) three times the other, (ii) it times the other, (iii) more than the other by h.

Solution:

(i) Let the roots of given eqn. be ax² + bx + c = 0 are α and 3α.

which gives the required condition.

(ii) Let the roots of given eqn. are α and nα

which gives the required condition.

(iii) Let the roots of given eqn. ax² + bx + c = 0 are α and α + h.

which gives the required condition.

Question 2.

Find the condition that the ratio between the roots of the equation ax² + bx + c = 0 may be m : n.

Solution:

Given quadratic eqn.be ax² + bx + c = 0 …(1)

Let the roots of eqn. (1) are mα and nα.

which gives the required condition

Question 3.

If the ratio of the roots of the equation x² + px + q = 0 is equal to the ratio of the roots of x² + lx + m = 0, prove that mp² = ql².

Solution:

Let α and ß are the roots of eqn. x² + px + q = 0

∴ α + ß = – p ; αß = q

and γ, δ are the roots of the eqn.

x² + lx + m = 0

γ + δ = – l and γδ = m

according to given condition, we have

which is the required condition.

Question 4.

For what values of a and b, the equation x² + (2a – 3) x = 3b + 4 should have both the roots zero ?

Solution:

Given quadratic eqn. be, x² + (2a – 3) x – (3b + 4) = 0 …(1)

Now both the roots of eqn. (1) are zero.

∴ coeff. of x = 0 and constant term = 0

⇒ 2a – 3 = 0 and 3b + 4 = 0 ⇒ a = \(\frac { 3 }{ 2 }\) and b = – \(\frac { 4 }{ 3 }\)

Question 5.

Find the values of X and p if both the roots of the equation (3λ + 1)x² = (2λ + 3λ) x – 3 are infinite.

Solution:

We know that if a and p are the roots of eqn. ax² + bx + c = 0.

Then \(\frac { 1 }{ α }\) and \(\frac { 1 }{ ß }\) are the roots of eqn. cx² + bx + a = 0

Hence both roots of ax² + bx + c = 0 are infinite if roots of cx² + bx + a = 0 are zero.

∴ a = 0 and b = 0

Thus both the roots of eqn. (3λ + 1) x² – (2λ + 3μ) x + 3 = 0 are infinite.

if coeff. of x² = 0 and coeff. of x = 0

if 3λ + 1 = 0 and 2λ + 3μ = 0

if λ = – \(\frac { 1 }{ 3 }\) and – \(\frac { 2 }{ 3 }\) + 3μ = 0 ⇒ μ = \(\frac { 2 }{ 9 }\)

Thus λ = – \(\frac { 1 }{ 3 }\) and μ = \(\frac { 2 }{ 9 }\)

Question 6.

Find m so that the roots of the equation \(\frac{x^2-b x}{a x-c}=\frac{m-1}{m+1}\) may be equal in magnitude and opposite in sign.

Solution:

The given quadratic equation be

Question 7.

(i) The roots of the quadratic equation 4x² – (5a + 1) x + 5a = 0, are p and q. If q = 1 + p, calculate the possible values of a, p and q.

(ii) Find the values of p for which the quadratic equation x² – px + p + 3 = 0 has

(a) coincident roots, (b) real distinct roots, (c) one positive and one negative root.

Solution:

(i) Given the roots of eqn. 4x² – (5a + 1) x + 5a = 0 are p and q.

(ii) Given quadratic eqn. be, x² – px + p + 3 = 0 …(1)

On comparing eqn. (1) with ax² + bx + c = 0

We have, a = 1 ; b = – p ; c = p + 3

(a) Roots of eqn. (1) are coincident roots

∴ discriminant = 0

⇒ b² – 4 ac = 0

⇒ (- p)² – 4 x 1 x (p + 3) = 0

⇒ p² – 4p – 12 > 0

⇒ (p + 2) (p – 6) = 0

⇒ p = – 2, 6

(b) roots of eqn. (1) are real and distinct.

∴ D > 0

⇒ b² – 4ac > 0

⇒ p² – 4p – 12 > 0

⇒ (p + 2) (p – 6) > 0

⇒ p < – 2 or p > 6

(c) given one root of eqn. (1) be positive and other be negative i.e. roots of eqn. (1) are of opposite sign.

∴ product of roots < 0

⇒ \(\frac { c }{ a }\) < 0

∴ c and a are of opposite sign

Here a = 1 > 0

∴ c < 0 ⇒ p + 3 < 0 ⇒ p < – 3

Question 8.

Find the values of m for which the quadratic equation x² – m (2x – 8) – 15 = 0 has

(i) equal roots, (ii) both roots positive.

Solution:

The given quadratic eqn. be x² – m (2x – 8) – 15 = 0

⇒ x² – 2mx + 8m – 15 = 0 …(1)

On comparing eqn. (1) with ax² + bx + c = 0 ; we have,

a = 1 ; b = – 2m and c = 8m – 15

(i) Roots of eqn. (1) are equal.

∴ discriminant = 0 ⇒ b² – 4ac = 0 (- 2m)² – 4 x 1 x (8m – 15) = 0

⇒ 4m² – 32m + 60 = 0

⇒ m² – 8m + 15 = 0

⇒ (m – 3) (m – 5) = 0

⇒ m = 3, 5

(ii) both roots of eqn. (1) are positive

∴ Sum of roots > 0 ⇒ – \(\frac { b }{ a }\) > 0 ⇒ \(\frac { b }{ a }\) < 0

i.e. b and a are opposite sign.

Here a = 1 > 0

∴ b < 0 ⇒ – 2m < 0 ⇒ m > 0

and product of roots > 0 ⇒ \(\frac { c }{ a }\) > 0

∴ c and a are of same sign.

Here a = 1 > 0 ∴ c > 0

⇒ 8m – 15 > 0

⇒ m > \(\frac { 15 }{ 8 }\)

Therefore m > 0 and m > \(\frac { 15 }{ 8 }\).

Question 9.

If a + b + c = 0, prove that the roots of ax² + bx + c = 0 are rational. Hence, show that the roots of (p + q) x² – 2px + (p – q) = 0 are rational.

Solution:

Given quadratic equation be ax² + bx + c = 0 …(1)

Clearly x = 1 satisfy eqn. (1)

∵ a + b + c = 0 (given)

Thus x = 1 be the roots of eqn. (1).

Since irrational or imaginary/non-real roots occurs in conjugate pair. Other root of eqn. (1) must be rational.

Hence the general condition for the roots of eqn. (1) are rational be a + b + c = 0

Also, the given quadratic eqn. be, (p + q) x² – 2px + p – q = 0 …(2)

Here a + b + c = p + q – 2p + p – q = 0

∴ roots of eqn. (2) are also rational.

Question 10.

Show that the roots of (x – b) (x – c) + (x – c) (x – a) + (x – a) (x – b) = 0 are real, and that they cannot be equal unless a = b = c.

Solution:

Given quadratic eqn. be (x – b) (x – c) + (x – c) (x – a) + (x – a) (x – b) = 0

⇒ 3x² – 2 (a + b + c) x + be + ca + ab = 0 …(1)

On comparing with Ax² + Bx + C = 0

We have, A = 3; B = – 2 (a + b + c) and c = ab + bc + ca

Now roots of eqn. (1) are real if D ≥ 0

Here D = b² – 4AC = [- 2 (a + b + c)]² – 4 x 3 (ab + bc + ca)

= 4 [(a + b + c)² – 3 (ab + bc + ca)]

= 4 [a² + b² + c² – ab – bc – ca]

= 2 [2a² + 2b² + 2c² – 2ab – 2bc – 2ca]

= 2 [(a – b)² + (b – c)² + (c – a)²] > 0 [∵ (a – b)² > 0 ; (b – c)² > 0 and (c – a)² > 0]

Hence the roots of given eqn. (1) are real.

Now the roots of given eqn. are equal

if Discriminant = 0 if (a – b)² + (b – c)² + (c – a)² = 0

i.e. if a – b = 0 and b – c = 0 and c – a = 0 i.e. if a = b = c

Hence the roots of eqn. (1) cannot be equal unless a = b = c

Question 11.

Determine the values of m for which the equations 3x² + 4mx + 2 = 0 and 2x² + 3x – 2 = 0 may have a common root.

Solution:

Let α be the common root of both given quadratic equations

3x² + 4mx + 2 = 0 and

2x² + 3x – 2 = 0

i.e. 3α² + 4mα + 2 = 0

2α² + 3α – 2 = 0

By cross multiplication method, we have

Question 12.

Find the value of k, so that the equation 2x² + kx – 5 = 0 and x² – 3x – 4 = 0 may have one root common.

Solution:

Let a be the common root of both given equations

2x² + kx – 5 = 0 and x² – 3x – 4 = 0

∴ 2a² + kα – 5 = 0 …(1)

α² – 3α – 4 = 0 …(2)

By cross multiplication method, we have

Question 13.

If ax² + bx + c = 0 and bx² + cx + a = 0 have a common root, prove that a + b + c = 0 or a = b = c.

Solution:

Let a be the common root of both equations ax² + bx + c = 0 and bx² + cx + a = 0

i.e. aα² + bα + c = 0 …(1)

bα² + cα + a = 0 …(2)

By cross multiplication method, we have

Question 14.

The equations x² + x + a = 0 and x² + ax + 1 = 0 have a common real root

(a) for no value of a

(b) for exactly one value of a

(c) for exactly two values of a

(d) for exactly three values of a

Solution:

Let α be the common real root of equations

x² + x + a = 0 and x² + ax + 1 = 0

∴ α² + α + a = 0 …(1)

and α² + aα + 1 = 0 …(2)

eqn. (1) – eqn. (2) gives ;

(1 – a)α + a – 1 = 0 ⇒ α = 1

∴ from (1); 1 + 1 + a = 0 ⇒ a = – 2