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