Find the roots of the following quadratic equations by factorisation :

Solution:

(i) $x^{2}-3 x-10=0$

$\Rightarrow x^{2}-5 x+2 x-10=0$

$\Rightarrow x(x-5)+2(x-5)=0$

$\Rightarrow(x+2)(x-5)=0$

$\Rightarrow x+2=0$ or $x-5=0$

$\Rightarrow x=-2$ or $x=5$

Hence, the two roots are $-2$ and 5 .

(ii) $2 x^{2}+x-6=0$

$\Rightarrow 2 x^{2}+4 x-3 x-6=0$

$\Rightarrow 2 x(x+2)-3(x+2)=0$

$\Rightarrow(x+2)(2 x-3)=0$

$\Rightarrow x+2=0$ or $2 x-3=0$

$\Rightarrow x=-2$ or $x=\frac{\mathbf{3}}{\mathbf{2}}$

(iii) $\sqrt{2} \mathbf{x}^{2}+\mathbf{7 x}+\mathbf{5} \sqrt{\mathbf{2}}=\mathbf{0}$

$\Rightarrow \sqrt{2} x^{2}+5 x+2 x+5 \sqrt{2}=0$

$\Rightarrow x(\sqrt{2} x+5)+\sqrt{2}(\sqrt{2} x+5)=0$

$\Rightarrow(x+\sqrt{2})(\sqrt{2} x+5)=0$

$\Rightarrow x=-\sqrt{2}$ or $-\frac{5}{\sqrt{2}}$

Hence, the two roots are $-\sqrt{\mathbf{2}}$ and $-\frac{\mathbf{5}}{\sqrt{\mathbf{2}}}$

(iv) $2 x^{2}-x+\frac{1}{8}=0$

or $16 x^{2}-8 x+1=0$

or $\quad(4 x-1)^{2}=0$

$\Rightarrow$ Both roots are given by $4 x-1=0$

i.e., $x=\frac{1}{4} .$ Hence, the roots are $\frac{1}{4}, \frac{1}{4}$

(v) $100 x^{2}-20 x+1=0$

$\Rightarrow 100 x^{2}-10 x-10 x+1=0$

$\Rightarrow 10 x(10 x-1)-1(10 x-1)=0$

$\Rightarrow(10 x-1)^{2}=0$

$\Rightarrow(10 x-1)=0 \quad$ or $(10 x-1)=0$

$\Rightarrow x=\frac{1}{10}$ or $x=\frac{1}{10}$