Question

Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) $2x^2-7x+3=0$
(ii) $2x^2+x-4=0$
(iii) $4x^2+4\sqrt{3x}+3=0$
(iv) $2x^2+x+4=0$

Answer 1

i)

$2x^2-7x+3=0\Rightarrow x^2-\frac{7x}{2}+\frac{3}{2}=0\Rightarrow x^2-\frac{7x}{2}+\frac{3}{2}+\frac{49}{16}=\frac{49}{16}\Rightarrow x^2-\frac{7x}{2}+\frac{49}{16}=\frac{49}{16}-\frac{3}{2}\Rightarrow {(x-\frac{7}{4})}^2=\frac{25}{16}\Rightarrow (x-\frac{7}{4})=\pm \frac{5}{4}\Rightarrow x=\frac{7}{4}+\frac{5}{4}\text{and}x=\frac{7}{4}-\frac{5}{4}\Rightarrow x=3,\frac{1}{2}$

ii)

$2x^2+x-4=0\Rightarrow x^2+\frac{x}{2}-2=0\Rightarrow x^2+\frac{x}{2}+\frac{1}{16}=2+\frac{1}{16}\Rightarrow {(x+\frac{1}{4})}^2=\frac{33}{16}\Rightarrow x+\frac{1}{4}=\pm \frac{\sqrt{33}}{4}\Rightarrow x=\frac{\sqrt{33}-1}{4},\frac{-\sqrt{33}-1}{4}$

iii)

$4x^2+4\sqrt{3}x+3=0\Rightarrow x^2+\sqrt{3}x+\frac{3}{4}=0\Rightarrow x^2+\sqrt{3}x+\frac{3}{4}=-\frac{3}{4}+\frac{3}{4}\Rightarrow {(x+\frac{\sqrt{3}}{2})}^2=0\Rightarrow x=-\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2}$

iv)

$2x^2+x+4=0$

$\Rightarrow x^2+\frac{x}{2}+2=0\Rightarrow x^2+\frac{x}{2}+2=0\Rightarrow x^2+\frac{x}{2}+\frac{1}{16}=\frac{1}{16}-2\Rightarrow {(x+\frac{1}{4})}^2=-\frac{31}{16}\Rightarrow x=-\frac{1}{4}\pm \frac{\sqrt{3}1i}{4}$