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# S Chand ICSE Maths Solutions Class 10 Chapter 5 Quadratic Equations Exercise 5D

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**Quadratic Equations***.*

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## 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) x^{2} – 4x + 3 = 0

(ii) x^{2} – 4x + 5 = 0

(iii) x^{2} – 4x + 4 = 0

(iv) x^{2} – x = 7

**S Chand ICSE Maths Solutions 1(i): **

Given that x^{2} – 4x + 3 = 0. Then, we have a = 1, b = -4, c = 3

Thus, the discriminant of the given quadratic equation is

D = b^{2 }– 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 x^{2} – 4x + 5 = 0. Then, we have a = 1, b = -4, c = 5

Thus, the discriminant of the given quadratic equation is

D = b^{2 }– 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 x^{2} – 4x + 4 = 0. Then, we have a = 1, b = -4, c = 4

Thus, the discriminant of the given quadratic equation is

D = b^{2 }– 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 x^{2} – x = 7 ⇒ x^{2} – x – 7 = 0. Then, we have a = 1, b = -1, c = -7

Thus, the discriminant of the given quadratic equation is

D = b^{2 }– 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**.

**S Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5D: Ques No 2**

Without finding the roots, comment on the nature of the roots of the following quadratic equations.

(i) 3x^{2} – 6x + 5 = 0

(ii) 5y^{2} + 12y – 9 = 0

(iii) x^{2} – 5x – 7 = 0

(iv) a^{2}x^{2} + abx = b^{2}, a ≠ 0.

(v) 9a^{2}b^{2}x^{2} – 48abcdx + 64c^{2}d^{2 }= 0, a ≠ 0, b ≠ 0.

(vi) 4x^{2} = 1

(vii) 64x^{2} – 112x + 49 = 0

(viii) 8x^{2} + 5x + 1 = 0

**S Chand ICSE Maths Solutions 2(i):**

Given that 3x^{2} – 6x + 5 = 0. Then, we have a = 3, b = -6, c = 5

Thus, the discriminant of the given quadratic equation is

D = b^{2 }– 4ac = (-6)^{2} – 4(3)(5) = 36 – 60 = -24

⇒ D < 0.

Therefore, the roots of the given quadratic equation are **Imaginary.**

**S Chand ICSE Maths Solutions 2(ii):**

Given that 5y^{2} + 12y – 9 = 0. Then, we have a = 5, b = 12, c = -9.

Thus, the discriminant of the given quadratic equation is

D = b^{2 }– 4ac = (12)^{2} – 4(5)(-9) = 144 + 180 = 324 = 18^{2}.

⇒ D > 0 and a Perfect Square.

Therefore, the roots of the given quadratic equation are **Real, Rational **and** Distinct**.

**S Chand ICSE Maths Solutions 2(iii):**

Given that x^{2} – 5x – 7 = 0. Then, we have a = 1, b = -5, c = -7.

Thus, the discriminant of the given quadratic equation is

D = b^{2 }– 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**.

**S Chand ICSE Maths Solutions 2(iv):**

Given that a^{2}x^{2} + abx = b^{2 }⇒ a^{2}x^{2} + abx – b^{2} = 0, A ≠ 0. Then, we have A = a^{2}, B = ab, C = -b^{2}.

Thus, the discriminant of the given quadratic equation is

D = b^{2 }– 4ac = (ab)^{2} – 4(a^{2})(-b^{2}) = 5a^{2}b^{2}.

⇒ D > 0 and NOT a Perfect Square.

Therefore, the roots of the given quadratic equation are **Real, Irrational **and** Distinct**.

**S Chand ICSE Maths Solutions 2(v):**

Given that 9a^{2}b^{2}x^{2} – 48abcdx + 64c^{2}d^{2 }= 0, a ≠ 0, b ≠ 0. Then, we have A = 9a^{2}b^{2}, B = – 48abcd, C = 64c^{2}d^{2}.

Thus, the discriminant of the given quadratic equation is

D = b^{2 }– 4ac = (-48abcd)^{2} – 4(9a^{2}b^{2})(64c^{2}d^{2}) = 0.

⇒ D = 0.

Therefore, the roots of the given quadratic equation are **Real **and** Equal**.

**S Chand ICSE Maths Solutions 2(vi):**

Given that 4x^{2} = 1 ⇒ 4x^{2} – 1 = 0. Then, we have a = 4, b = 0, c = -1.

Thus, the discriminant of the given quadratic equation is

D = b^{2 }– 4ac = (0)^{2} – 4(4)(-1) = 0 + 16 = 4^{2}.

⇒ D > 0 and a Perfect Square.

Therefore, the roots of the given quadratic equation are **Real, Rational **and** Distinct**.

**S Chand ICSE Maths Solutions 2(vii):**

Given that 64x^{2} – 112x + 49 = 0. Then, we have a = 64, b = -112, c = 49.

Thus, the discriminant of the given quadratic equation is

D = b^{2 }– 4ac = (-112)^{2} – 4(64)(49) = 12544 – 12544 = 0.

⇒ D = 0.

Therefore, the roots of the given quadratic equation are **Real **and** Equal**.

**S Chand ICSE Maths Solutions 2(viii):**

Given that 8x^{2} + 5x + 1 = 0. Then, we have a = 8, b = 5, c = 1.

Thus, the discriminant of the given quadratic equation is

D = b^{2 }– 4ac = (5)^{2} – 4(8)(1) = 25 – 32 = -7.

⇒ D < 0.

Therefore, the roots of the given quadratic equation are **Imaginary**.

**S Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5D: Ques No 3**

Find the value of k so that the equation has equal roots (or coincident roots).

(i) 4x^{2} + kx + 9 = 0

(ii) kx^{2} – 5x + k = 0

(iii) 9x^{2} + 3kx + 4 = 0

(iv) x^{2} + 7(3 + 2k) – 2x(1 + 3k) = 0.

(v) (k – 12)x^{2} + 2(k – 12)x + 2^{ }= 0

**S Chand ICSE Maths Solutions 3(i):**

Given that 4x^{2} + kx + 9 = 0. Then, we have a = 4, b = k, c = 9.

Thus, the discriminant of the given quadratic equation is

D = b^{2 }– 4ac = (k)^{2} – 4(4)(9) = k^{2} – 144.

Since the roots of the given quadratic equation are equal.

⇒ D = 0.

⇒ k^{2} – 144 = 0

⇒ k = ±√144

⇒ k = ±12

**S Chand ICSE Maths Solutions 3(ii):**

Given that kx^{2} – 5x + k = 0. Then, we have a = k, b = -5, c = k.

Thus, the discriminant of the given quadratic equation is

D = b^{2 }– 4ac = (-5)^{2} – 4(k)(k) = 25 – 4k^{2}.

Since the roots of the given quadratic equation are equal.

⇒ D = 0.

⇒ 25 – 4k^{2} = 0

⇒ k = ±√(25/4)

⇒ k = ±5/2

**S Chand ICSE Maths Solutions 3(iii):**

Given that 9x^{2} + 3kx + 4 = 0. Then, we have a = 9, b = 3k, c = 4.

Thus, the discriminant of the given quadratic equation is

D = b^{2 }– 4ac = (3k)^{2} – 4(9)(4) = 9k^{2 }– 144

Since the roots of the given quadratic equation are equal.

⇒ D = 0.

⇒ 9k^{2 }– 144 = 0

⇒ k = ±√(144/9)

⇒ k = ±√16

⇒ k = ±4

**S Chand ICSE Maths Solutions 3(iv):**

Given that x^{2} + 7(3 + 2k) – 2x(1 + 3k) = 0 ⇒ x^{2} – 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 = b^{2 }– 4ac

= (-2(1 + 3k))^{2} – 4(1)(7(3 + 2k))

= 4(1 + 6k + 9k^{2}) – 28(3 + 2k)

= 4 + 24k + 36k^{2} – 84 – 56k

= 36k^{2} – 32k – 80 = 4(9k^{2} – 8k – 20)

Since the roots of the given quadratic equation are equal.

⇒ D = 0.

⇒ 4(9k^{2} – 8k – 20) = 0

⇒ (9k^{2} – 18k + 10k – 20) = 0

⇒ 9k(k – 2) + 10(k – 2) = 0

⇒ (9k + 10)(k – 2) = 0

⇒ k = 2 or -10/9

**S Chand ICSE Maths Solutions 3(v):**

Given that (k – 12)x^{2} + 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 = b^{2 }– 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

**S Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5D: Ques No 4**

For what value of k, do the following quadratic equations have real roots.

(i) 2x^{2} + 2x + k = 0

(ii) kx^{2} – 2x + 2 = 0

(iii) 2x^{2} – 10x + k = 0

(iv) kx^{2} + 8x – 2 = 0

(v) 9x^{2} – 24x + k^{ }= 0

(vi) x^{2} – 4x + k^{ }= 0

**S Chand ICSE Maths Solutions 4(i):**

Given that 2x^{2} + 2x + k = 0. Then, we have a = 2, b = 2, c = k.

Thus, the discriminant of the given quadratic equation is

D = b^{2 }– 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

**S Chand ICSE Maths Solutions 4(ii):**

Given that kx^{2} – 2x + 2 = 0. Then, we have a = k, b = –2, c = 2.

Thus, the discriminant of the given quadratic equation is

D = b^{2 }– 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

**S Chand ICSE Maths Solutions 4(iii):**

Given that 2x^{2} – 10x + k = 0. Then, we have a = 2, b = –10, c = k.

Thus, the discriminant of the given quadratic equation is

D = b^{2 }– 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

**S Chand ICSE Maths Solutions 4(iv):**

Given that kx^{2} + 8x – 2 = 0. Then, we have a = k, b = 8, c = -2.

Thus, the discriminant of the given quadratic equation is

D = b^{2 }– 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

**S Chand ICSE Maths Solutions 4(v):**

Given that 9x^{2} – 24x + k^{ }= 0. Then, we have a = 9, b = -24, c = k.

Thus, the discriminant of the given quadratic equation is

D = b^{2 }– 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

**S Chand ICSE Maths Solutions 4(vi):**

Given that x^{2} – 4x + k^{ }= 0. Then, we have a = 1, b = -4, c = k.

Thus, the discriminant of the given quadratic equation is

D = b^{2 }– 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

**S Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5D: Ques No 5**

Find the value of k for which the given equation has equal roots. Also, find the roots.

(i) 9x^{2} – 24x + k = 0

(ii) 2kx^{2} – 40x + 25 = 0

**S Chand ICSE Maths Solutions 5(i):**

Given that 9x^{2} – 24x + k = 0. Then, we have a = 9, b = -24, c = k.

Thus, the discriminant of the given quadratic equation is

D = b^{2 }– 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.

**S Chand ICSE Maths Solutions 5(ii):**

Given that 2kx^{2} – 40x + 25 = 0. Then, we have a = 2k, b = -40, c = 25.

Thus, the discriminant of the given quadratic equation is

D = b^{2 }– 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.

**S Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5D: Ques No 6**

Find the value of k for which the given equation has real and distinct roots.

(i) kx^{2} + 2x + 1 = 0

(ii) kx^{2} + 6x + 1 = 0

**S Chand ICSE Maths Solutions 6(i):**

Given that kx^{2} + 2x + 1 = 0. Then, we have a = k, b = 2, c = 1.

Thus, the discriminant of the given quadratic equation is

D = b^{2 }– 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

**S Chand ICSE Maths Solutions 6(i):**

Given that kx^{2} + 6x + 1 = 0. Then, we have a = k, b = 6, c = 1.

Thus, the discriminant of the given quadratic equation is

D = b^{2 }– 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

**S Chand ICSE Maths Solutions Class 10 Quadratic Equations Exercise 5D: Ques No 7**

If -5 is a root of the quadratic equation 2x^{2} + px – 15 = 0 and the quadratic equation p(x^{2} + x) + k = 0 has equal roots, find the value of k.

**S Chand ICSE Maths Solutions:**

Since -5 is a root of the quadratic equation 2x^{2} + 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(x^{2} + x) + k = 0

⇒ 7(x^{2} + x) + k = 0

⇒ 7x^{2} + 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)x^{2} + (c – a)x + (a – b) = 0 are equal, then prove that 2b = a + c.

**S Chand ICSE Maths Solutions: **

Given that (b – c)x^{2} + (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