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 \(x = \frac{1}{10 }\)