S Chand ICSE Maths Solutions Class 10 Chapter 5 Quadratic Equations Exercise 5D
Hi students, Welcome to Amans Maths Blogs (AMB). In this post, you will get S Chand ICSE Maths Solutions Class 10 Chapter 5 Quadratic Equations Exercise 5D. This is the chapter of introduction of ‘Quadratic Equations‘ included in ICSE 2021 Class 10 Maths syllabus.
S Chand ICSE Maths Solutions Class 10 Chapter 5 Quadratic Equations Exercise 5D
S Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5D: Ques No 1
Use the discriminant to determine the nature of the roots of each of the following quadratic equations.
(i) x2 – 4x + 3 = 0
(ii) x2 – 4x + 5 = 0
(iii) x2 – 4x + 4 = 0
(iv) x2 – x = 7
S Chand ICSE Maths Solutions 1(i):
Given that x2 – 4x + 3 = 0. Then, we have a = 1, b = -4, c = 3
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac = (-4)2 – 4(1)(3) = 16 – 12 = 4 = (2)2
⇒ D > 0 and Perfect Square.
Therefore, the roots of the given quadratic equation are Real, Rational and Distinct.
S Chand ICSE Maths Solutions 1(ii):
Given that x2 – 4x + 5 = 0. Then, we have a = 1, b = -4, c = 5
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac = (-4)2 – 4(1)(5) = 16 – 20 = -4
⇒ D < 0.
Therefore, the roots of the given quadratic equation are Imaginary.
S Chand ICSE Maths Solutions 1(iii):
Given that x2 – 4x + 4 = 0. Then, we have a = 1, b = -4, c = 4
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac = (-4)2 – 4(1)(4) = 16 – 16 = 0
⇒ D = 0.
Therefore, the roots of the given quadratic equation are Real and Equal.
S Chand ICSE Maths Solutions 1(iv):
Given that x2 – x = 7 ⇒ x2 – x – 7 = 0. Then, we have a = 1, b = -1, c = -7
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac = (-1)2 – 4(1)(-7) = 1 + 28 = 29
⇒ D > 0 and NOT a Perfect square.
Therefore, the roots of the given quadratic equation are Real and Irrational.
Ques No 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
Solutions 2(i):
Given that 3x2 – 6x + 5 = 0. Then, we have a = 3, b = -6, c = 5
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac = (-6)2 – 4(3)(5) = 36 – 60 = -24
⇒ D < 0.
Therefore, the roots of the given quadratic equation are Imaginary.
Solutions 2(ii):
Given that 5y2 + 12y – 9 = 0. Then, we have a = 5, b = 12, c = -9.
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac = (12)2 – 4(5)(-9) = 144 + 180 = 324 = 182.
⇒ D > 0 and a Perfect Square.
Therefore, the roots of the given quadratic equation are Real, Rational and Distinct.
Solutions 2(iii):
Given that x2 – 5x – 7 = 0. Then, we have a = 1, b = -5, c = -7.
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac = (-5)2 – 4(1)(-7) = 25 + 28 = 53
⇒ D > 0 and NOT a Perfect Square.
Therefore, the roots of the given quadratic equation are Real, Irrational and Distinct.
Solutions 2(iv):
Given that a2x2 + abx = b2 ⇒ a2x2 + abx – b2 = 0, A ≠ 0. Then, we have A = a2, B = ab, C = -b2.
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac = (ab)2 – 4(a2)(-b2) = 5a2b2.
⇒ D > 0 and NOT a Perfect Square.
Therefore, the roots of the given quadratic equation are Real, Irrational and Distinct.
Solutions 2(v):
Given that 9a2b2x2 – 48abcdx + 64c2d2 = 0, a ≠ 0, b ≠ 0. Then, we have A = 9a2b2, B = – 48abcd, C = 64c2d2.
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac = (-48abcd)2 – 4(9a2b2)(64c2d2) = 0.
⇒ D = 0.
Therefore, the roots of the given quadratic equation are Real and Equal.
Solutions 2(vi):
Given that 4x2 = 1 ⇒ 4x2 – 1 = 0. Then, we have a = 4, b = 0, c = -1.
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac = (0)2 – 4(4)(-1) = 0 + 16 = 42.
⇒ D > 0 and a Perfect Square.
Therefore, the roots of the given quadratic equation are Real, Rational and Distinct.
Solutions 2(vii):
Given that 64x2 – 112x + 49 = 0. Then, we have a = 64, b = -112, c = 49.
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac = (-112)2 – 4(64)(49) = 12544 – 12544 = 0.
⇒ D = 0.
Therefore, the roots of the given quadratic equation are Real and Equal.
Solutions 2(viii):
Given that 8x2 + 5x + 1 = 0. Then, we have a = 8, b = 5, c = 1.
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac = (5)2 – 4(8)(1) = 25 – 32 = -7.
⇒ D < 0.
Therefore, the roots of the given quadratic equation are Imaginary.
Ques No 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
Solutions 3(i):
Given that 4x2 + kx + 9 = 0. Then, we have a = 4, b = k, c = 9.
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac = (k)2 – 4(4)(9) = k2 – 144.
Since the roots of the given quadratic equation are equal.
⇒ D = 0.
⇒ k2 – 144 = 0
⇒ k = ±√144
⇒ k = ±12
Solutions 3(ii):
Given that kx2 – 5x + k = 0. Then, we have a = k, b = -5, c = k.
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac = (-5)2 – 4(k)(k) = 25 – 4k2.
Since the roots of the given quadratic equation are equal.
⇒ D = 0.
⇒ 25 – 4k2 = 0
⇒ k = ±√(25/4)
⇒ k = ±5/2
Solutions 3(iii):
Given that 9x2 + 3kx + 4 = 0. Then, we have a = 9, b = 3k, c = 4.
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac = (3k)2 – 4(9)(4) = 9k2 – 144
Since the roots of the given quadratic equation are equal.
⇒ D = 0.
⇒ 9k2 – 144 = 0
⇒ k = ±√(144/9)
⇒ k = ±√16
⇒ k = ±4
Solutions 3(iv):
Given that x2 + 7(3 + 2k) – 2x(1 + 3k) = 0 ⇒ x2 – 2(1 + 3k)x + 7(3 + 2k) = 0. Then, we have a = 1, b = – 2(1 + 3k), c = 7(3 + 2k).
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac
= (-2(1 + 3k))2 – 4(1)(7(3 + 2k))
= 4(1 + 6k + 9k2) – 28(3 + 2k)
= 4 + 24k + 36k2 – 84 – 56k
= 36k2 – 32k – 80 = 4(9k2 – 8k – 20)
Since the roots of the given quadratic equation are equal.
⇒ D = 0.
⇒ 4(9k2 – 8k – 20) = 0
⇒ (9k2 – 18k + 10k – 20) = 0
⇒ 9k(k – 2) + 10(k – 2) = 0
⇒ (9k + 10)(k – 2) = 0
⇒ k = 2 or -10/9
Solutions 3(v):
Given that (k – 12)x2 + 2(k – 12)x + 2 = 0. Then, we have a = 1, b = 2(k – 12), c = 2(k – 12).
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac
= (2(k – 12))2 – 4(1)(2(k – 12))
= 4(k – 12)2 – 8(k – 12)
= 4(k – 12)[k – 12 – 2]
= 4(k – 12)(k – 14)
Since the roots of the given quadratic equation are equal.
⇒ D = 0.
⇒ 4(k – 12)(k – 14) = 0
⇒ k = 12 or 14
Ques No 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
Solutions 4(i):
Given that 2x2 + 2x + k = 0. Then, we have a = 2, b = 2, c = k.
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac
= (2)2 – 4(2)(k)
= 4 – 8k
Since the roots of the given quadratic equation are real.
⇒ D ≥ 0.
⇒ 4 – 8k ≥ 0
⇒ -4(2k – 1) ≥ 0
⇒ k – 1/2 ≤ 0
⇒ k ≤ 1/2
Solutions 4(ii):
Given that kx2 – 2x + 2 = 0. Then, we have a = k, b = –2, c = 2.
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac
= (-2)2 – 4(k)(2)
= 4 – 8k
Since the roots of the given quadratic equation are real.
⇒ D ≥ 0.
⇒ 4 – 8k ≥ 0
⇒ -4(2k – 1) ≥ 0
⇒ k – 1/2 ≤ 0
⇒ k ≤ 1/2
Solutions 4(iii):
Given that 2x2 – 10x + k = 0. Then, we have a = 2, b = –10, c = k.
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac
= (-10)2 – 4(2)(k)
= 100 – 8k
Since the roots of the given quadratic equation are real.
⇒ D ≥ 0.
⇒ 100 – 8k ≥ 0
⇒ -4(2k – 25) ≥ 0
⇒ 2k – 25 ≤ 0
⇒ k ≤ 25/2
Solutions 4(iv):
Given that kx2 + 8x – 2 = 0. Then, we have a = k, b = 8, c = -2.
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac
= (8)2 – 4(k)(-2)
= 64 + 8k
= 8(8 + k)
Since the roots of the given quadratic equation are real.
⇒ D ≥ 0.
⇒ 8(8 + k) ≥ 0
⇒ (8 + k) ≥ 0
⇒ k + 8 ≥ 0
⇒ k ≥ -8
Solutions 4(v):
Given that 9x2 – 24x + k = 0. Then, we have a = 9, b = -24, c = k.
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac
= (-24)2 – 4(9)(k)
= 576 – 36k
= -36(k – 16)
Since the roots of the given quadratic equation are real.
⇒ D ≥ 0.
⇒ -36(k – 16) ≥ 0
⇒ (k – 16) ≤ 0
⇒ k ≤ 16
Solutions 4(vi):
Given that x2 – 4x + k = 0. Then, we have a = 1, b = -4, c = k.
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac
= (-4)2 – 4(1)(k)
= 16 – 4k
= -4(k – 4)
Since the roots of the given quadratic equation are real.
⇒ D ≥ 0.
⇒ -4(k – 4) ≥ 0
⇒ (k – 4) ≤ 0
⇒ k ≤ 4
Ques No 5
Find the value of k for which the given equation has equal roots. Also, find the roots.
(i) 9x2 – 24x + k = 0
(ii) 2kx2 – 40x + 25 = 0
Solutions 5(i):
Given that 9x2 – 24x + k = 0. Then, we have a = 9, b = -24, c = k.
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac
= (-24)2 – 4(9)(k)
= 576 – 36k
= -36(k – 16)
Since the roots of the given quadratic equation are equal.
⇒ D = 0.
⇒ -36(k – 16) = 0
⇒ k = 16
And, Root will be α = β = -b/2a = -(-24)/2(9) = 4/3.
Solutions 5(ii):
Given that 2kx2 – 40x + 25 = 0. Then, we have a = 2k, b = -40, c = 25.
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac
= (-40)2 – 4(2k)(25)
= 1600 – 200k
= -200(k – 8)
Since the roots of the given quadratic equation are equal.
⇒ D = 0.
⇒ -200(k – 8) = 0
⇒ k = 8
And, Root will be α = β = -b/2a = -(-40)/2(2k) = 10/k = 10/8 = 5/4.
Ques No 6
Find the value of k for which the given equation has real and distinct roots.
(i) kx2 + 2x + 1 = 0
(ii) kx2 + 6x + 1 = 0
Solutions 6(i):
Given that kx2 + 2x + 1 = 0. Then, we have a = k, b = 2, c = 1.
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac
= (2)2 – 4(k)(1)
= 4 – 4k
= -4(k – 1)
Since the roots of the given quadratic equation are real and distinct roots.
⇒ D > 0.
⇒ -4(k – 1) > 0
⇒ (k – 1) < 0
⇒ k < 1
Solutions 6(ii):
Given that kx2 + 6x + 1 = 0. Then, we have a = k, b = 6, c = 1.
Thus, the discriminant of the given quadratic equation is
D = b2 – 4ac
= (6)2 – 4(k)(1)
= 36 – 4k
= -4(k – 9)
Since the roots of the given quadratic equation are real and distinct roots.
⇒ D > 0.
⇒ -4(k – 9) > 0
⇒ (k – 9) < 0
⇒ k < 9
Ques No 7
If -5 is a root 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.
Solutions:
Since -5 is a root of the quadratic equation 2x2 + px – 15 = 0, then x = -5 will satisfy the equation.
Thus,
2(-5)2 + p(-5) – 15 = 0
⇒ 50 – 5p – 15 = 0
⇒ 35 – 5p = 0
⇒ p = 35/5 = 7
Now, put p = 7 in p(x2 + x) + k = 0
⇒ 7(x2 + x) + k = 0
⇒ 7x2 + 7x + k = 0
This equation has equal roots. Then, its discriminant is D = 0.
⇒ (7)2 – 4(7)(k) = 0
⇒ k = 7/4
S Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5D: Ques No 8
If the roots of the equation (b – c)x2 + (c – a)x + (a – b) = 0 are equal, then prove that 2b = a + c.
S Chand ICSE Maths Solutions:
Given that (b – c)x2 + (c – a)x + (a – b) = 0. Then, we have A = (b – c), B = (c – a), C = (a – b).
At x = 1, (b – c) + (c – a) + (a – b) = 0. Thus, α = 1.
Since the roots of the given quadratic equation are equal. Thus, α = β = 1.
⇒ αβ = 1
⇒ (a – b)/(b – c) = 1
⇒ a – b = b – c
⇒ 2b = a + c
| S Chand ICSE Class 10 Maths Chapterwise Notes & Solutions |
|---|
| Chapter 1 : GST Notes & Solutions | Notes | Exercise 1 |
| Chapter 2 : Banking Notes & Solutions | Notes | Exercise 2 | Revision Exercise |
| Chapter 3 : Shares & Dividends Notes & Solutions | Notes | Exercise 3A | Exercise 3B | Revision Exercise |
| Chapter 4 : Linear Inequations in One Variable Notes & Solutions | Notes | Exercise 4 | Revision Exercise |
| Chapter 5 : Quadratic Equations Notes & Solutions | Notes | Exercise 5A | Exercise 5B | Exercise 5C| Exercise 5D| Exercise 5E| Revision Exercise |
| Chapter 6 : Ratios & Proportions Notes & Solutions | Notes | Exercise 6A | Exercise 6B | Exercise 6C| Revision Exercise |
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