Students can cross-reference their work with OP Malhotra Class 10 Solutions Chapter 5 Quadratic Equations Ex 5(d) to ensure accuracy.
S Chand Class 10 ICSE Maths Solutions Chapter 5 Quadratic Equations Ex 5(d)
Question 1.
Use the discriminant to determine the nature of the roots of each of the following quadratic equations :
(1) x² – 4x + 3 = 0
(ii) x² – 4x + 5 – 0
(iii) x² – 4x + 4 = 0
(iv) x² – x = 7
Solution:
(i) x² – 4x + 3 = 0
Here a= 1, b = – 4, c = 3
∴ D = b² – 4ac = (- 4)² – 4 x 1 x 3 = 16 – 12 = 4
∵ D > 0,
∴Roots are real and distinct and rational.
(ii) x² – 4x + 5 = 0
Here a = 1, b = – 4, c = 5
∴ D = b² – 4ac = (- 4)² – 4 x 1 x 5 = 16 – 20 = – 4
∵ D < 0
∴ Roots are not real or roots are imaginary.
(iii) x² – 4x + 4 = 0
Here a = 1, b = – 4, c = 4
∴ D = b² – 4ac = (- 4)² – 4 x 1 x 4
= 16 – 16 = 0
∵ D = 0
∴ Roots are real and equal
(iv) x² – x = 7 ⇒ x² – x – 7 = 0
Here a = 1, b = – 1, c = – 7
∴ D = b² – 4ac = (- 1)² – 4 x 1 x (- 7)
= 1 + 28 = 29
∵ D > 0
∴ Roots are real and distinct (irrational).
Question 2.
Without finding the roots, comment on the nature of the roots of the following quadratic equations.
(i) 3x² – 6x + 5 = 0
(ii) 5y2 + 12y – 9 = 0
(iii) x² – 5x – 7 = 0
(iv) a²x² + abx = b², a ≠ 0
(v) 9a²b²x² – 48abcdx + 64c²d² = 0, a ≠ 0, b ≠ 0
(vi) 4x² = 1
(vii) 64x² – 112x + 49 = 0
(viii) 8x² + 5x + 1 = 0
Solution:
(i) 3x² – 6x + 5 = 0
Here a = 3, b = – 6, c = 5 D = b² – 4ac = (- 6)² – 4 x 3 x 5
= 36 – 60 = – 24
∵ D < 0
∴ Roots are not real i.e. roots are imaginary.
(ii) 5y² + 12y – 9 = 0
Here a = 5, b = 12, c = – 9
∴ D = b² – 4ac = (12)² – 4 x 5 x (- 9)
= 144 + 180 = 224
∵ D > 0
∴ Roots are real, distinct and rational.
(iii) x² – 5x – 7 = 0
Here a = 1, b = – 5, c = – 7
∴ D = b² – 4ac = (- 5)² – 4 x 1 x (- 7)
= 25 + 28 = 53
∵ D > 0
∴ Roots are real, distinct and rational.
(iv) a²x² + abx = b²
⇒ a²x² + abx – b² = 0
Here A = a², B = ab. C = – b²
∴ D – b² – 4AC = (ab)² – 4 x a² (-b²)
= a²b² + 4a²b² = 5 a²b²
∵ D > 0
∴ Roots are real, distinct and rational.
(v) 9a²b²x² – 48abcdx + 64c²cd² = 0, a ≠ 0, b ≠ 0
Here A = 9a²b², B = – 48abed, C = 64c²d²
∴ D = b² – 4AC = (- 48abcd)² – 4 x 9a²b² x 64c²d²
= 2304a²b²c²d² – 2304a²b²c²d² = 0
∵ D = 0
∴ Roots are real equal.
(vi) 4x² = 1 ⇒ 4x² – 1 = 0
Here a = 4, b = 0, c = – 1
∴ D = b² – 4ac = (-112)² – 4 x 4 x (- 1)
= 0+ 16 = 16 = (4)²
∵ D > 0
∴ Roots are real, distinct.
(vii) 64x² – 112x + 49 = 0
Here a = 64, b = – 112, c = 49
∴D = b² – 4ac = (- 112)² – 4 x 64 x 49
= 12544 – 12544 = 0
∵ D = 0
∴ Roots are real and equal.
(viii) 8x² + 5x + 1 = 0
Here a = 8, b = 5, c = 1
∴ D = b² – 4ac = (5)² – 4 x 8 x 1
= 25 – 32 = – 7
∵ D < 0
∴ Roots are not real, but are imaginary.
Question 3.
Find the value of ‘k’ so that the equation has equal roots (or coincident roots)
(i) 4x² + kx + 9 = 0
(ii) kx² – 5x + k = 0
(iii) 9x² + 3kx + 4 = 0
(iv) x² + 7 (3 + 2k) – 2x (1 + 3k) = 0
(v) (k – 12) x² + 2 (k – 12) x + 2 = 0
Solution:
(i) 4x² + kx + 9 = 0 Here a = 4, b = k and c = 9
∴ D = b² – 4ac = a² – 4 x 4 x 9 = k² – 144
∵ Roots are equal
∴ D = 0
⇒ k² – 144 = 0
⇒ (k)² – (12)² = 0
⇒ (k + 12) (k – 12) = 0
Either k + 12 = 0, then k = – 12
or k – 12 = 0, then k = 12
∴ k = 12, – 12
(ii) kx² – 5x + k = 0
Here a = k, b = – 5, c = k
∴ D = b² – 4ac = (- 5)² – 4 x k x k
= 25 – 4k²
∵ Roots are equal
∴D = 0
⇒ 25 – 4k² = 0 ⇒ 4k² – 25 = 0
⇒ (2k)² – (5)² = 0
⇒ (2k + 5) (2k – 5) = 0
Either 2k + 5 = 0, then 2k = – 5 ⇒ k = \(\frac { – 5 }{ 2 }\)
or 2k – 5 = 0, then 2A = 5 ⇒ k = \(\frac { 5 }{ 2 }\)
(iii) 9x² + 3kx + 4 = 0
Here a = 9, b = 3k, c = 4
∴ D = a² – 4ac = (3k)² – 4 x 9 x 4
= 9k² – 144
∵ Roots are equal
∴D = 0 ⇒ 9k² – 144 = 0 ⇒ k² – 16 = 0 (Dividing by 9)
⇒ (k)² – (4)² = 0 ⇒ (k + 4) (k – 4) = 0
Either k + 4 = 0, then k = – 4 or k – 4 = 0, then k = 4
k = 4, – 4
(iv) x² + 7 (3 + 2k) – 2x (1 + 3k) = 0
⇒ x² – 2 (1 + 3k)x + 7 (3 + 2k) = 0
Here a = 1, b = – 2 (1 + 3k), c = 7 (3 + 2k)
∴ D = a² – 4ac
= [- 2 (1 + 3k)]² – 4 x 1 x 7 (3 + 2k)
= 4(1+ 9k² + 6k) – 28 (3 + 2k)
= 4 + 36k² + 24k – 84 – 56k
= 36k ² – 32k – 80
∵ Roots are equal
∴ D = 0
⇒ 36k² – 32k – 80 = 0
⇒ 9k² – 8k – 20 = 0 (Dividing by 4)
⇒ 9k² – 18k + 10k – 20 = 0
⇒ 9k(k – 2)+ 10(k – 2) = 0
⇒ (k – 2) (9k + 10) = 0
Either k – 2 = 0, then k = 2
or 9k+ 10 = 0, then 9k = – 10 ⇒ k = \(\frac { -10 }{ 9 }\)
∴ k = 2, \(\frac { -10 }{ 9 }\)
(v) (k – 12) x² + 2 (k – 12) x + 2 = 0
Here a = k – 12, A = 2 (A – 12), c = 2
∴ D = a² – 4ac = [2 (k – 12)]² – 4 x (k – 12) x 2
= 4(k – 12)² – 8 (k – 12)
∵ Roots are equal
∴ D = 0
⇒ 4 (k – 12)² – 8 (k – 12) = 0
⇒ (k – 12)² – 2 (k – 12) = 0 (Dividing by 4)
(k – 12) (k – 12 – 2) = 0
⇒ (k – 12) (k – 14) = 0
Either k – 12 = 0, then k = 12
or k – 14 = 0, then k = 14
∴ k = 12, 14
Question 4.
For what value of k, do the following quadratic equations have real roots.
(i) 2x² + 2x + k = 0
(ii) kx² – 2x + 2 = 0
(iii) 2x² – 10x + k = 0
(iv) kx² + 8x – 2 = 0
(v) 9x² – 24x + k = 0
(vi) x² – 4x + k = 0
Solution:
(i) 2x² + 2x + k = 0
Here a = 2, b = 2, c = k
∴ D = b² – 4ac = (2)² – 4 x 2 x k
= 4 – 8k
∵ Roots are real
∴ D > 0 ⇒ 4 – 8k ≥ 0
⇒ 1 – 2x ≥ 0 (Dividing by 4)
⇒ 1 ≥ 2x ⇒ 2x ≤ 1
⇒ k ≤ \(\frac { 1 }{ 2 }\)
(ii) kx² – 2x + 2 = 0
Here a = k, b = – 2, c = 2
∴ D = b² – 4ac = (- 2)² – 4 x k x 2
= 4 – 8k
∵ Roots are real
⇒ D ≥ 0 ⇒ 4 – 8k ≥ 0
⇒ 1 – 2x ≥ 0 (Dividing by 4)
⇒ 1 ≥ 2k ⇒ 2k ≤ 1
⇒ k ≤ \(\frac { 1 }{ 2 }\)
(iii) 2x² – 10x + k = 0
Here a = 2, b = – 10, c = k
∴ D = b² – 4ac = (-10)² – 4 x 2 x k
= 100 – 8k
∵ Roots are real
∴ D ≥ 0 ⇒ 100 – 8x ≥ 0
⇒ 25 – 2k ≥ 0 (Dividing by 4)
⇒ 25 ≥ 2x ⇒ 2k ≤ 25
⇒ k ≤ \(\frac { 25 }{ 2 }\)
(iv) kx² + 8x – 2 = 0
Here a = k, b = 8, c = – 2
∴ D = b² – 4ac
= (8)² – 4 x k x (- 2) = 64 + 8k
∵ Roots are real
∴ D ≥ 0 ⇒ 64 + 8k = 0
⇒ 8 + k ≥ 0
⇒ k ≥ – 8
∴ k ≥ – 8
(v) 9x² – 24x + k = 0
Here a = k, b = – 24, c = k
∴ D = b² – 4ac – (- 24)² – 4 x 9 x k
= 576 – 36k
∵ Roots are real
∴ D ≥ 0 ⇒ 576 – 36k ≥ 0
⇒ 16 – x ≥ 0 (Dividing by 36)
⇒ 16 ≥ k ⇒ x ≤ 16
(vi) x² – 4x + x = 0
Here a = 1, b = – 4, c = x
∴ D = b² – 4ac = (- 4)² – 4 x 1 x k
= 16 – 4k
∵ Roots are real
∴ D ≥ 0 ⇒ 16 – 4k ≥ 0
⇒ 4 – k ≥ 0 (Dividing by 4)
⇒ 4 ≥ k ⇒ k ≤ 4
∴ k ≤ 4
Question 5.
Find the value of x for which the given equation has equal roots. Also find the roots :
(i) 9x4 – 24x + k = 0
(ii) 2kx² – 40x + 25
Solution:
(i) 9x² – 24x + k = 0
Here a = 9, b = – 24, c = k
∴ D = b² – 4ac = (- 24)² – 4 x 9 x k
= 576 – 36x v
∵ Roots are equal
∴ D = 0
⇒ 576 – 36k = 0 ⇒ 36k = 576
⇒ k = \(\frac { 576 }{ 36 }\)
∴ x = 16
Now 9x² – 24 + 16 = 0
⇒ (3x)² – 2 x 3x × 4 + (4)² = 0
⇒ (3x – 4)² = 0 ⇒ 3x – 4 = 0
∴ 3x = 4 ⇒ x = \(\frac { 4 }{ 3 }\)
Hence x = \(\frac { 4 }{ 3 }\)
(ii) 2kx² – 40x + 25 = 0
Here a = 2k, b = – 40, c = 25
∴ D = b² – 4ac = (- 40)² – 4 x 2k x 25
= 1600 – 200x
∵ Roots are equal
∴ D = 0 ⇒ 1600 – 200k = 0
⇒ 200k = 1600 ⇒ k = \(\frac { 1600 }{ 200 }\) = 8
∴ k = 8
Now 2 x 8x² – 40x + 25 = 0
⇒ 16x² – 40x + 25 = 0
⇒ (4x)² – 2 x 4x × 5 + (5)² = 0
⇒ (4x – 5)² = 0
⇒ 4x = 5 ⇒ x = 5
⇒ 4x = 5 ⇒ x = \(\frac { 5 }{ 4 }\)
∴ k = 8, x = \(\frac { 5 }{ 4 }\)
Question 6.
Find the values of k for which the given quadratic equations has real and distinct roots :
(i) kx² + 2x + 1 = 0
(ii) kx² + 6x + 1 = 0
Solution:
(i) kx² + 2x + 1 = 0
Here a = k, b = 2, c = 1
∴ D = b² – 4ac = (2)² – 4 x k x 1
= 4 – 4k
∵ Roots are real and distinct
∴ D > 0 ⇒ 4 – 4k > 0
⇒ 4 > 4k ⇒ 1 > k (Dividing by 4)
⇒ x < 1
∴ k < 1
(ii) kx² + 6x + 1 = 0
Here a = k, b = 6, c – 1
∴ D = b² – 4ac = (6)² – 4 x k < 1 = 36 – 4k ∵ Roots are real and distinct ∴ D > 0 ⇒ 36 – 4x > 0
⇒ 36 > 4k ⇒ 9 > k (Dividing by 4)
⇒ k < 9
∴ k < 9
Question 7.
If – 5 is a roots of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots, find the value of k.
Solution:
∵ – 5 is a root of 2x² + px – 15 = 0
∴ – 5 will satisfy it
∴ 2 (- 5)² + p (- 5) – 15 = 0
⇒ 2 x 25 – 5p – 15 = 0
⇒ 50 – 5p – 15 = 0
⇒ 35 – 5p = 0
⇒ 5p = 35 ⇒ p = \(\frac { 35 }{ 5 }\) = 7
Now ∵ (x² + x) + x = 0 has equal roots
⇒ px² + px + x = 0
⇒ 7x² + 7x + x = 6 (∵p = 7)
Here a = 7, b = 7, c = k
∴ D = b² – 4ac = (7)² – 4 x k = 49 – 28 k
∵ Roots are equal
∴ D = 0 ⇒ 49 – 28k = 0
⇒ 28x = 49 ⇒ k = \(\frac { 49 }{ 28 }\) = \(\frac { 7 }{ 4 }\)
∴ k = \(\frac { 7 }{ 4 }\)
Question 8.
If the roots of the equation (b – c) x² + (c – a) x + (a – b) = 0 are equal, then prove that 2b = a + c.
Solution:
∵ (b – c) x² + (c – a) x + (a – b) = 0 has equal roots
∴ D = 0
Now here A = b – c, B = c – a and C = a – b
D = B² – 4AC = (c – a)² – 4x (b – c) (a – b)
∴ D = 0
∴ (c – a)² – 4 (b – c) (a – b) = 0
⇒ c² + a² – 2ac – 4 (ab – b² – ca + bc) = 0
⇒ c² + a² – 2ac – 4ab + 4b² + 4ca – 4bc = 0
a² + 4b² + c² – 4ab – 4bc + 2ca = 0
⇒ (a)² + (2b)² + (c)² – 2 x a x 2b – 2 x 2b x c + 2c x a = 0
⇒ (a – 2b + c)² = 0 ⇒ a – 2b + c = 0
⇒ a + c = 2b ⇒ 2b = a + c
Hence proved.