Question

Solve :

i) $3x^2-2\sqrt{6}x+2=0$
ii) $2\sqrt{3}{x}^2-5x+\sqrt{3}=0$
iii) $x^2+x+2=0$

Answer 1

(i) $3x^2-2\sqrt{6}x+2=0$ $3\times 2=6$

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

$\Rightarrow \sqrt{3}x[\sqrt{3}x-\sqrt{2}]-\sqrt{2}[\sqrt{3}x-\sqrt{2}]=0$

$\Rightarrow (\sqrt{3}x-\sqrt{2}][\sqrt{3}x-\sqrt{2}]=0$

$\Rightarrow (\sqrt{3}x-\sqrt{2}{)}^2=0$

$\Rightarrow x=\frac{\sqrt{2}}{\sqrt{3}}$

$\Rightarrow x=\sqrt{\frac{2}{3}}$

(ii) $2\sqrt{3}{x}^2-5x+\sqrt{3}=0$ $2\sqrt{3}\times \sqrt{3}=6$

$\Rightarrow 2\sqrt{3}{x}^2-2x-3x+\sqrt{3}=0$ $6\times 1=6$ $2\times 3=6$

$\Rightarrow 2x[\sqrt{3}x-1]-\sqrt{3}[\sqrt{3}x-1]=0$ $6-1=5$ $2+3=5$

$\Rightarrow [\sqrt{3}x-1][2x-\sqrt{3}]=0$

$\Rightarrow x=\frac{\sqrt{3}}{2}$ or $x=\frac{1}{\sqrt{3}}$.

(iii) $x^2+x+2=0$ $2\times 1=2$

$b^2-4ac$

$\Rightarrow 1^2-4\left(2\right)\Rightarrow -7<0$

$\therefore x^2+x+2=0$ has no real roots as $b^2-4ac<0$.