Question

Find the roots of the following quadratic equations by factorisation:
(i) $x^2 – 3x – 10 = 0$

(ii) $2x^2 + x – 6 = 0$

(iii) $\sqrt2x^2+7x+5\sqrt2 = 0$

(iv) 2x^2 – x + \frac{1}8$$ = 0

(v) $100x^2 – 20x + 1 = 0$

Answer 1

(i) $x^2 – 3x – 10 = 0$

= $x^2 -5x +2x -10$
= $x(x-5)+ 2 (x-5)$
= $(x-5)(x+2)$

Roots of this equation are the values for which (x-5)(x+2) =0
∴ x-5=0 or x+2 = 0

i.e., x = 5 or x = −2

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(ii) $2x^2 + x – 6 = 0$

= $2x^2 +4x -3x -6$
= $2x (x+2) -3 (x+2)$
= $(x+2)(2x-3)$

Roots of this equation are the values for which $(x+2)(2x-3) =0$
∴$x+2 =0$ or $2x-3 = 0$

i.e., $x = -2$ or $x = \frac{3}2$

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(iii) $\sqrt2x^2+7x+5\sqrt2 = 0$

=$\sqrt2x^2 +5x+2x +5\sqrt2$
= $x(\sqrt2x+5) + \sqrt2(\sqrt2x +5)$
=$(\sqrt2x+5)(x+\sqrt2)$

Roots of this equation are the values for which$(\sqrt2x+5)(x+\sqrt2) =0$
∴ $\sqrt2x+5 =0$ or x+\sqrt2$$= 0

i.e., x = $-\frac{5}{\sqrt2 }$or x = $-\sqrt2$

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(iv) $2x^2 – x + \frac{1}8 = 0$

=$\frac{1}8(16x^2 -8x +1)$
= $\frac{1}8 (16x^2 -4x -4x +1)$
= $\frac{1}{8} (4x(4x-1) -1(4x-1))$
= $\frac{1}8 (4x-1)^2$

Roots of this equation are the values for which $(4x-1)(4x-1) =0$
∴ $4x-1=0$ or $4x-1 = 0$

i.e., $x = \frac{1}{4}$ or $x = \frac{1}{4}$

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(v) $100x^2 – 20x + 1 = 0$

= $100x^2 -10x -10x +1$
= $10x(10x-1) -1 (10x-1)$
= $(10x-1)^2$

Roots of this equation are the values for which $(10x-1)(10x-1) =0$
∴ $10x-1=0$ or $10x-1 = 0$

i.e., $x = \frac{1}{10 }$ or $x = \frac{1}{10 }$