OP Malhotra Class 11 Maths Solutions Chapter 10 Quadratic Equations Ex 10(c)

Utilizing OP Malhotra Maths Class 11 Solutions Chapter 10 Quadratic Equations Ex 10(c) as a study aid can enhance exam preparation.

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

Question 1.
Without solving, find the nature of the roots of the following equations :
(i) 3x² – 7x + 5 = 0.
(ii) 4x² + 4x + 1 = 0.
(iii) 3x² + 7x + 2 = 0.
(iv) x² + px – q² = 0.
Solution:
(i) Given quadratic eqn. be 3x² – 7x + 5 = 0
On comparing with ax² + bx + c = 0 ; we have
a = 3; b = – 7; c = 5
Here discriminant D = b² – 4ac
= (- 7)² – 4 x 3 x 5
= 49 – 60 = – 11 < 0
Hence the roots of given quadratic eqn. are imaginary.

(ii) Given quadratic eqn. be 4x² + 4x + 1 = 0
On comparing with ax² + bx + c = 0, we have a = 4; b = 4; c = 1
Here discriminant D = b² – 4ac
= 16 – 4 x 4 x 1 = 0
∴ roots are real and equal.

(iii) Given quadratic eqn. be 3x² + 7x + 2 = 0
On comparing with ax² + bx + c = 0, we have a = 3 ; b = 7 ; c = 2
Here discriminant D = b² – 4ac
= 7² – 4 x 3 x 2
= 49 – 24
= 25 > 0
∴ roots are real, distinct and rational

(iv) Given quadratic eqn. be
x² + px – q² = 0 …(1)
On comparing eqn. (1) with
ax² + bx + c = 0,
we have a = 1 ; b = p ; c = – q²
Here discriminant D = b² – 4ac
= p² – 4 x 1 x (- q²)
= p² + 4q² > 0
Hence the roots are real and unequal if p and q both not equal to 0 and roots are real and equal if p = 0 and q = 0

Question 2.
If the equation
(1 + m²) x² + 2mcx + c² – a² = 0 has equal roots, show that c² = a² (1 + m²).
Solution:
Given quadratic eqn. be,
(1 + m²)x² + 2mcx + c² – a² = 0 …(1)
On comparing with Ax² + Bx + C = 0
we have, A = 1 + m²; B = 2mc
and C = c² – a²
given eqn. (1) has equal roots.
∴ b² – 4AC = 0
⇒ 4m²c² – 4 (1 + m²) (c² – a²) = 0
⇒ 4m²c² – 4 {c² – a² + m²c² – a²m²} = 0
⇒ – 4 (c² – a² – a²m²) = 0
⇒ c² = a²(1 + m²)

Question 3.
Find the value of m so that the roots of the equation (4 – m)x² + (2m + 4) x + (8m + 1) = 0 may be equal.
Solution:
Given quadratic equation be,
(4 – m)x² + (2m + 4)x + 8m + 1 = 0
On comparing with ax² + bx + c = 0 ; we have
a = 4 – m ; b = 2m + 4 and c = 8m + 1
Since it is given that roots of eqn. (1) are equal
∴ discriminant = 0 ⇒ b² – 4ac = 0
⇒ (2m + 4)² – 4 x (4 – m) (8m + 1) = 0
⇒ 4m² + 16m + 16 – 4 (32m + 4 – 8m² – m) = 0
⇒ 36m² – 108m = 0
⇒ 36m (m – 3) = 0
⇒ m = 0, 3

Question 4.
If the roots of ax² + x + b = 0be real and unequal, show that the roots of \(\frac{x^2+1}{x}=4 \sqrt{a b}\) are imaginary.
Solution:
Given roots of ax² + x + b = 0 be real and unequal.
∴ discriminant > 0
⇒ 1 – 4ab > 0 …(1)
Also, given quadratic eqn. can be written as,
x² – 4\(\sqrt{ab}\) x + 1 = 0 …(2)
On comparing eqn. (2) with Ax² + Bx + C = 0
we have, A = 1 ; B = – 4\(\sqrt{ab}\) ; C = 1
Here discriminant = B² – 4AC
= (- 4\(\sqrt{ab}\))² – 4 x 1 x 1
= 16ab – 4 = 4 (4ab – 1)
= – 4(1 – 4ab) < 0 [using eqn. (1)]
Hence the roots of eqn. (2) are imaginary.

Question 5.
Find a so that the sum of the roots of the equation ax² + 2x + 3a = 0 may be equal to their product.
Solution:
Given quadratic eqn. be
ax² + 2x + 3a = 0 … (1)
On comparing eqn. (1) with
Ax² + Bx + C = 0
we have, A = a; B = 2; C = 3a
∴ Sum of roots = – \(\frac { B }{ A }\) = – \(\frac { 2 }{ a }\)
product of roots = \(\frac { C }{ A }\) = \(\frac { 3a }{ a }\) = 3
It is given that, sum of roots = product of roots
⇒ \(\frac { – 2 }{ a }\) = 3 ⇒ a = \(\frac { -2 }{ 3 }\)

Question 6.
If α, ß are the roots of the equation x² + x + 1 = 0, find the value of α³ + ß³.
Solution:
Since α, ß are the roots of eqn. x² + x + 1 = 0
∴ S = α + ß = \(\frac { – 1 }{ 1 }\) = – 1
and P = αß = \(\frac { 1 }{ 1 }\) = 1
∴ α³ + ß³ = (α + ß)³ – 3αß (α + ß)
= (- 1)³ – 3 (1)(- 1) = – 1 + 3 = 2

Question 7.
If α, ß are the roots of the equation x² + px + q = 0, find the value of
(a) α³ß + αß³
(b) α⁴ + α²ß² + ß⁴.
Solution:
Given α and ß are the roots of equation x² + px + q = 0
∴ α + ß = – p ; ap = q
(a) α³ß + αß³ = αß (α² + ß²)
= αß [(α + ß)² – 2αß]
= q (- P)² – 2q] = q(p² – 2q)

(b) b⁴ + α²ß² + ß⁴ = (α² + ß²)² – α²ß²
= [(α + ß)² – 2αß]² – (αß )²
= [(- P)² – 2q]² – q²
= (p² – 2q)² – q²
= p⁴ – 4p²q + 3q²

Question 8.
If the roots of the equation x² + px + 7 = 0 are denoted by α and ß, and α² + ß² = 22, find the possible values of p.
Solution:
Since it is given that a and p are the roots of equation x² + px + 7 = 0
∴ α + ß = – p ; αß = 7
Also given α² + ß² = 22
⇒ (α + ß)² – 2αß = 22
⇒ (- p)² – 2 x 7 = 22
⇒ p² = 22+ 14 = 36
⇒ p = ± 6

Question 9.
If α, ß are the roots of the equation 3x² – 6x + 4 = 0, find the value of
\(\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)+2\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)+3 \alpha \beta\)
Solution:
Given a and p are the roots of equation 3x² – 6x + 4 = 0

Question 10.
If α, ß are the roots of ax² + bx + c = 0, find the values of
(i) \(\left(\frac{\alpha}{\beta}-\frac{\beta}{\alpha}\right)^2\)
(ii) \(\frac{\alpha^3}{\beta}+\frac{\beta^3}{\alpha}\)
Solution:
Given a and p are the roots of eqn. ax² + bx + c = 0

Question 11.
If the sum of the roots of the equation x² – px + q = 0 be m times their difference, prove that p² (m² – 1) = 4m²p.
Solution:
Let α, ß are the roots of equation
x² – px + q = 0 …(1)
∴ α + ß = p ; αß = q
It is given that sum of roots of eqn. (1)
= m x difference of their roots
⇒ α + ß = m|α – ß|
On squaring both sides ; we have
(α + ß)² = m² (α – ß)²
⇒ (α + ß)² = m² [(α + ß)² – 4αß]
⇒ p² = m² [p² – 4q]
⇒ p² (m² – 1) = 4m²q

Question 12.
If one root of the equation x² + ax + 8 = 0 is 4 while the equation x² + ax + b = 0 has equal roots, find b.
Solution:
Since 4 be the root of equation
x² + ax + 8 = 0 ⇒ 4² + 4a + 8 = 0
⇒ 4a + 24 = 0 ⇒ a = – 6
Since the equation x² + ax + b = 0 has equal roots
∴ Discriminant = 0
⇒ α² – 4b = 0 ⇒ (- 6)² – 4b = 0
⇒ 4b = 36 ⇒ b = 9

Question 13.
Find the value of a for which one root of the quadratic equation (a² – 5a + 3) x² + (3a – 1) x + 2 = 0 is twice as large as the other.
Solution:
Let α and 2α are the two roots of given equation (a² – 5a + 3)x² + (3a – 1)x + 2 = 0

Question 14.
If α, ß are the roots of the equation ax² – bx + b = 0, prove that
\(\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}-\sqrt{\frac{b}{a}}\) = 0.
Solution:
Given α and ß are the roots of the equation ax² – bx + b = 0
∴ α + ß = \(\frac { b }{ a }\) and αß = \(\frac { b }{ a }\)
Now, \(\frac { 1 }{ 2 }\)
= \(\frac { 1 }{ 2 }\)

Question 15.
If α and ß are the roots of the equation x² + x – 7 = 0, form the equation whose roots are α² and ß².
Solution:
Given a and P are the roots of the eqn. x² + x – 7 = 0
∴ α + ß = – 1; αß = – 7
Sum of roots of required eqn. = α² + ß² = S
= (α + ß)² – 2αß = (- 1)² – 2 (- 7)
= 15
and product of roots of required equation
= α² p² = (αß)² = (- 7)² = 49
Thus the required quadratic eqn. by given by x² – Sx + P = 0 ⇒ x² – 15x + 49 = 0

Question 16.
If α and ß are the roots of the equation 2x² + 3x + 2 = 0, find the equation whose roots are α + 1 and ß + 1.
Solution:
Given α and ß are the roots of the eqn.
2x² + 3x + 2 = 0
∴ α + ß = – \(\frac { 3 }{ 2 }\) ; αß = \(\frac { 2 }{ 2 }\) = 1
∴ S = sum of roots of quadratic eqn.
= α + 1 + ß + 1
= – \(\frac { 3 }{ 2 }\) + 2 = \(\frac { 1 }{ 2 }\)
P = product of roots of quadratic eqn.
= (α + 1)(α + 1)
= αß + (α + ß) + 1
= 1 – \(\frac { 3 }{ 2 }\) + 1 = \(\frac { 1 }{ 2 }\)
Hence the required quadratic eqn. be given by x² – Sx + P = 0 ⇒ x² – \(\frac { 1 }{ 2 }\)x + \(\frac { 1 }{ 2 }\) = 0
⇒ 2x² – x + 1 = 0

Question 17.
Find the equation whose roots are \(\frac { α }{ ß }\) and \(\frac { ß }{ α }\), where α and ß are the roots of the equation x² + 2x + 3 = 0.
Solution:
Given a and p are the roots of equation x² + 2x + 3 = 0
∴ α + ß = – 2 ; αß = 3
S = sum of roots of required eqn.
= \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha \beta}\)
= \(\frac{(\alpha+\beta)^2-2 \alpha \beta}{\alpha \beta}=\frac{(-2)^2-2 \times 3}{3}=-\frac{2}{3}\)
P = product of roots of required eqn.
= \(\frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha}\) = 1
Hence the required quadratic eqn. having roots \(\frac { α }{ ß }\) and \(\frac { ß }{ α }\) is given by
x² – Sx + P = 0 ⇒ x² – (- \(\frac { 2 }{ 3 }\)) x + 1 = 0
⇒ 3x² + 2x + 3 = 0

Question 18.
If a and p are the roots of the equation 2x² – 3x + 1 = 0, form the equation whose roots are \(\frac{\alpha}{2 \beta+3} \text { and } \frac{\beta}{2 \alpha+3}\)
Solution:
Given a and p are the roots of the eqn. 2x² – 3x + 1 = 0

Question 19.
If a ≠ b and a² = 5a – 3, b² = 5b – 3, then form that equation whose roots are \(\frac { a }{ b }\) and \(\frac { b }{ a }\).
Solution:
Given a ≠ b and a² = 5a – 3 and b² = 5b – 3
∴ a and b are roots of the quadratic equation x² – 5x + 3 = 0
∴ a + b = 5 ; ab = 3
S = sum of roots of required eqn.
= \(\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{a b}\)
= \(\frac{(a+b)^2-2 a b}{a b}=\frac{5^2-2 \times 3}{3}=\frac{19}{3}\)
and P = product of roots of required eqn.
= \(\frac{a}{b} \times \frac{b}{a}\) = 1
Hence the required quadratic eqn. having roots \(\frac { a }{ b }\) and \(\frac { b }{ a }\) be given by
x² – Sx + P = 0
⇒ x² – x + 1 = 0 ⇒ 3x² – 19x + 3 = 0

Question 20.
Given that a and P are the roots of the equation x² = x + 7.
(i) Prove that
(a) \(\frac { 1 }{ α }\) = \(\frac{\alpha-1}{7}\) and (b) α³ = 8α + 7.
(ii) Find the numerical value of \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\).
Solution:
(i) Since α be the root of eqn.
x² = x + 7 …(1)
∴ it must satisfies eqn. (1)
∴ α² = α + 7 ⇒ α² – α = 7
⇒ α (α – 1) = 7 ⇒ \(\frac { 1 }{ α }\) = \(\frac{\alpha-1}{7}\)

(b) Further α³ = α² . α = (α + 7) α
= α² + 7α = α + 7 + 7α
⇒ α³ = 8α + 7

(ii) Since α, ß are the roots of eqn. (1)
∴ α + ß = 1 ; αß = – 7
∴ \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha \beta}\)
= \(\frac{(\alpha+\beta)^2-2 \alpha \beta}{\alpha \beta}\)
= \(\frac{1^2-2 \times(-7)}{-7}=\frac{-15}{7}\)

Question 21.
Given that α and ß are the roots of the equation x² – x + 7 = 0, find
(i) the numerical value of \(\frac{\alpha}{\beta+3}+\frac{\beta}{\alpha+3}\) ;
(ii) an equation whose roots are \(\frac { α }{ ß+3 }\) and \(\frac { ß }{ α +3 }\).
Solution:
Given α and ß are the roots of equation x² – x + 7 = 0
∴ α + ß = + 1 ; αß = 7
(i) S = sum of roots of required eqn.

(ii) Hence the required quadratic eqn. having roots \(\frac { α }{ ß+3 }\) and \(\frac { ß }{ α + 3 }\)
x² – Sx + P = 0
⇒ x² + \(\frac { 10 }{ 19 }\)x + \(\frac { 7 }{ 19 }\) = 0
⇒ 19x² + 10x + 7 = 0

Question 22.
Given that α and ß are the roots of the equation 2x² – 3x + 4 = 0, find an equation whose roots are α + \(\frac { 1 }{ α }\) and ß + \(\frac { 1 }{ ß }\).
Solution:
Given α and ß are the roots of the eqn.

Thus the required quadratic eqn. having roots α + \(\frac { 1 }{ α }\) and ß + \(\frac { 1 }{ ß }\) be given by
x² – Sx + p = 0 ⇒ x² – \(\frac { 9 }{ 4 }\)x + \(\frac { 13 }{ 8 }\) = 0
⇒ 8x² – 18x + 13 = 0

Question 23.
The roots of the quadratic equation x² + px + 8 = 0 are α and ß. Obtain the values of p, if
(i) α = ß²
(ii) α – ß = 2.
Solution:
Given α and ß are the roots of the eqn.
x² + px + 8 = 0
∴ α + ß = – p ; αß = 8 … (1)
(i) When α = ß² ∴ from (1); we have
ß³ = 8 ⇒ ß = 2 ∴ α = 2² = 4
∴ from (1) ; – p = 4 + 2 ⇒ p = – 6

(ii) Given α – ß = 2 ;
On squaring both sides ; (α – ß)² = 4
⇒ (α + ß)² – 4αß = 4
⇒ (- p)² – 4 x 8 = 4
⇒ p² = 36 ⇒ p = ± 6

Question 24.
If the roots of x² – bx + c = 0 be two consecutive integers, then find the value of b² – 4c.
Solution:
Let the roots of eqn.
x² – 6x + c = 0 are α, α + 1
∴ α + α + 1 = b ⇒ 2α + 1 = b …(1)
and α (α + 1) = c …(2)
Eliminating a from eqn. (1) and eqn. (2)
From (1); α = \(\frac { b – 1 }{ 2 }\)
∴ from (2); \(\left(\frac{b-1}{2}\right)\left(\frac{b-1}{2}+1\right)\) = c
⇒ \(\left(\frac{b-1}{2}\right)\left(\frac{b+1}{2}\right)\) = c
⇒ b² – 1 = 4c
⇒ b² – 4c = 1

Question 25.
The roots of the equation
px² – 2(p + 2)x + 3p = 0 are α and ß. If α – ß = 2, calculate the value of α, ß and p.
Solution:
Given a and p are the roots of the equation
px² – 2(p + 2)x + 3p = 0 = 0
∴ α + ß = \(\frac{2(p+2)}{p}\) ; αß = \(\frac { 3p }{ p }\) = 3
Also given α – ß = 2 ; on squaring both sides; we have

On solving eqn. (2) and (3); we have α = – 1 and ß = – 3

Question 26.
The roots of the equation ax² + bx + c = 0 are α and ß. Form the quadratic equation whose roots are α + \(\frac { 1 }{ ß }\) and ß + \(\frac { 1 }{ α }\).
Solution:
Given α and ß are the roots of the eqn. ax² + bx + c = 0

Thus the required quadratic eqn. having roots α + \(\frac { 1 }{ ß }\) and ß + \(\frac { 1 }{ α }\) is given by
x² – Sx + P = 0
⇒ x² + \(\frac{b(c+a) x}{a c}+\frac{(a+c)^2}{a c}\) = 0
⇒ acx² + b (c + a) x + (a + c)² = 0

Question 27.
Two candidates attempt to solve a quadratic equation of the form x² + px + q = 0. One starts with a wrong value of p and finds the roots to be 2 and 6. The other starts with a wrong value of q and finds the roots to be 2 and – 9. Find the correct roots and the equation.
Solution:
One candidate find the roots of given eqn.
x² + px + q = 0 …(1)
are 2 and 6 ∴ it forms the quadratic eqn. as
(x – 2) (x – 6) = 0
⇒ x² – 8x + 12 = 0 …(2)
∴ Correct value of q = 12 [on comparing eqn. (1) and eqn. (2)]
[since value of p is given to be wrong]
Other candidate find the roots of eqn. (1) are 2 and – 9 ∴ he forms the quadratic eqn. as : (x – 2) (x + 9) = 0
⇒ x² + 7x – 18 = 0 …(3)
So on comparing eqn. (1) and eqn. (3) ;
we have Correct value of p = 7
Hence the correct values of p and q are 7 and 12
∴ required eqn. becomes; x² + 7x + 12 = 0
⇒ (x + 3) (x + 4) = 0 ⇒ x = – 3, – 4
Hence the correct roots of given eqn. are – 3 and – 4.

Question 28.
Given that a and P are the roots of the equation x² = 7x + 4.
(i) Show that α³ = 53α + 28
(ii) find the value of \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\).
Solution:
Since α and ß are the roots of eqn. x² = 7x + 4
i. e. x² – 7x – 4 = 0 …(1)
∴ α + ß = 7 and αß = – 4
Since a be the root of eqn. (1) ∴ it satisfies eqn. (1)
∴ α² = 7α + 4
⇒ α³ = 7α² + 4α = 7 (7α + 4) + 4α = 53α + 28

(ii) \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha \beta}=\frac{(\alpha+\beta)^2-2 \alpha \beta}{\alpha \beta}\)
= \(\frac{7^2-2 \times(-4)}{-4}=\frac{57}{-4}\)

Question 29.
The ratio of the roots of the equation x² + αx + α + 2 = 0 is 2. Find the values of the parameter a.
Solution:
Since the ratio of the roots of the equation x² + αx + α + 2 = 0 is 2.
Let 2ß and ß are the roots of given equation.
∴ 2ß + ß = – a ⇒ 3ß = – α …(1)
and (2ß) ß = α + 2 ⇒ 2ß² = α + 2 …(2)
From (1) and (2); we have
\(2\left(-\frac{\alpha}{3}\right)^2=\alpha+2 \Rightarrow 2 \times \frac{\alpha^2}{9}=\alpha+2\)
⇒ 2α² – 9α – 18 = 0
∴ α = \(\frac{9 \pm \sqrt{81+4 \times 2 \times 18}}{2 \times 2}=\frac{9 \pm 15}{4}\)
∴ α = 6, – \(\frac { 3 }{ 2 }\)

Question 30.
If (1 – p) is a root of the quadratic equation x² + px + (1 – p) = 0, then its roots are
(a) 0, – 1
(b) – 1, 1
(c) 0, 1
(d) – 1, 2
Solution:
Since (1 – p) be the root of quadratic equation
x² + px + (1 – p) = 0 …(1)
⇒ (1 – p)² + p(1 – p) + (1 – p) = 0
⇒ (1 – P) [1 – p + p + 1] = 0
⇒ 2(1 – p) = 0
⇒ p = 1
∴ eqn. (1) becomes ; x² + x = 0
⇒ x (x + 1) = 0 ⇒ x = 0, – 1
Hence the required roots of given equation are 0 and – 1.