SChand CLASS 10 Chapter 5 Quadratic Equations Exercise 5D

Exercise 5D

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Q1 | Ex-5D | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper

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Question 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) $x^2 – x = 7$

Sol :

(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.

(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.

(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.

(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.

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Q2 | Ex-5D | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper

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Question 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) $a^2x^2 + abx = b^2$, a ≠ 0.

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

(vi) $4x^2 = 1$

(vii) $64x^2 – 112x + 49 = 0$

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

Sol :

(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.

(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.

(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.

(iv)

Given that $a^2x^2 + abx$

$= b^2$

⇒ $a^2x^2 + abx – b^2$= 0, A ≠ 0.

Then, we have $A = a^2$, B = ab, $C = -b^2$.

Thus, the discriminant of the given quadratic equation is

$D = b^2 – 4ac$

$= (ab)^2 – 4(a^2)(-b^2)$

$= 5a^2b^2$

⇒ D > 0 and NOT a Perfect Square.

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

(v)

Given that $9a^2b^2x^2 – 48abcdx + 64c^2d^2 = 0$, a ≠ 0, b ≠ 0.

Then, we have $A = 9a^2b^2$, B = – 48abcd, $C = 64c^2d^2$.

Thus, the discriminant of the given quadratic equation is

$D = b^2 – 4ac$

$= (-48abcd)^2 – 4(9a^2b^2)(64c^2d^2) = 0$.

⇒ D = 0.

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

(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.

(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.


(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.

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Q3 | Ex-5D | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper

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Question 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) $(k – 12)x^2 + 2(k – 12)x + 2 = 0$

Sol :

(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


(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

$D = b^2 – 4ac$

$= (-5)^2 – 4(k)(k)$

$= 25 – 4k2$

Since the roots of the given quadratic equation are equal.

⇒ D = 0.

⇒ $25 – 4k^2 = 0$

⇒ $k = ±\sqty{\frac{25}{4}}$

⇒ $k = ±\frac{5}{2}$


(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.

⇒ $9k^2 – 144 = 0$

⇒ $k = ±\sqrt{\frac{144}{9}}$

⇒ k = ±√16

⇒ k = ±4

 

(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 $\frac{-10}{9}$

(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

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Q4 | Ex-5D | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper

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Question 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) $x^2 – 4x + k = 0$

Sol :

(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

⇒ $k – \frac{1}{2}$ ≤ 0

⇒ k ≤ $\frac{1}{2}$

(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

⇒ $k – \frac{1}{2}$ ≤ 0

⇒ k ≤ $\frac{1}{2}$

(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 ≤ $\frac{25}{2}$

(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

(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

(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

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Q5 | Ex-5D | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper

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Question 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) $2kx^2 – 40x + 25 = 0$

Sol :

(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 $= \frac{-b}{2a} = \frac{-(-24)}{2(9)} = \frac{4}{3}$.

(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 $= \frac{-b}{2a} = \frac{-(-40)}{2(2k)} = \frac{10}{k} = \frac{10}{8}= \frac{5}{4}$.

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Q6 | Ex-5D | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper

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Question 6

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

(i) $kx^2 + 2x + 1 = 0$

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

Sol :

(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


(ii)

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

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Q7 | Ex-5D | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper

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Question 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.

Sol :

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 $= \frac{35}{5}$ = 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.

⇒ $(7)^2 – 4(7)(k) = 0$

⇒ $k = \frac{7}{4}$

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Q8 | Ex-5D | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper

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Question 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.

Sol :

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

⇒ $\frac{(a – b)}{(b – c)}= 1$

⇒ a – b = b – c

⇒ 2b = a + c