Question

Solve the following equations using factorization method.
(i) $x^2-3x-10=0$
(ii) $2x^2+x-6=0$
(iii) $\sqrt{2}{x}^2+7x+5\sqrt{2}=0$
(iv) $2x^2-x+\frac{1}{8}=0$
(v) $100x^2-20x+1=0$

Answer 1

$i)x^2-3x-10=0\Rightarrow x^2-5x+2x-10=0\Rightarrow x(x-5)+2(x-5)=0\Rightarrow (x-5\left)\right(x+2)\Rightarrow (x-5)=0\text{or}(x+2)=0\therefore x=5,-2ii)2x^2+x-6=0\Rightarrow 2x^2+4x-3x-6=0\Rightarrow 2x(x+2)-3(x+2)=0\Rightarrow (2x-3)(x+2)=0\Rightarrow (2x-3)=0\text{or}(x+2)=0\therefore x=\frac{3}{2},-2iii)\sqrt{2}{x}^2+7x+5\sqrt{2}=0\Rightarrow \sqrt{2}{x}^2+5x+2x+5\sqrt{2}=0\Rightarrow x(\sqrt{2}x+5)+\sqrt{2}(\sqrt{2}x+5)=0\Rightarrow (x+\sqrt{2}\left)\right(\sqrt{2}x+5)=0\Rightarrow (x+\sqrt{2})=0\text{or}(\sqrt{2}x+5)=0\Rightarrow x=-\sqrt{2},\frac{-5}{\sqrt{2}}iv)2x^2-x+\frac{1}{8}=0\Rightarrow 16x^2-8x+1=0\Rightarrow {\left(4x\right)}^2-2.4x+{\left(1\right)}^2=0\Rightarrow {(4x-1)}^2=0\Rightarrow x=\frac{1}{4}v)100x^2-20x+1=0\Rightarrow 100x^2-10x-10x+1=0\Rightarrow 10x(10x-1)-1(10x-1)=0\Rightarrow (10x-1\left)\right(10x-1)=0\Rightarrow {(10x-1)}^2=0\Rightarrow x=\frac{1}{10}$