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) $4x^2+4\sqrt{3}x+3=0$ (iv) $2x^2+x+4=0$

Answer 1

$a).$ $2x^2-7x+3=0$

$\Rightarrow x^2-\frac{7}{2}x=-\frac{3}{2}$

Adding ${\left(\frac{7}{4}\right)}^2$ on both sides

$\Rightarrow x^2-\frac{7}{2}x+{\left(\frac{7}{4}\right)}^2=\frac{-3}{2}+{\left(\frac{7}{4}\right)}^2$

$\Rightarrow {(x-\frac{7}{4})}^2=\frac{-3}{2}+\frac{49}{16}$

$\Rightarrow {(x-\frac{7}{4})}^2=\frac{25}{16}$

$\Rightarrow {(x-\frac{7}{4})}^2={\left(\frac{5}{4}\right)}^2$

Taking square root on both sides

$\Rightarrow (x-\frac{7}{4})=\pm \frac{5}{4}$

$\Rightarrow x-\frac{7}{4}=\frac{5}{4},$ $x-\frac{7}{4}=\frac{-5}{4}$

$x=\frac{5}{4}+\frac{7}{4}$ $x=\frac{-5}{4}+\frac{7}{4}$

$x=3$ $x=\frac{1}{2}$

$b).$ $2x^2+x-4=0$

$x^2+\frac{x}{2}=2$

Adding ${\left(\frac{1}{4}\right)}^2$ on both sides

$\Rightarrow x^2+\frac{x}{2}+{\left(\frac{1}{4}\right)}^2=2+{\left(\frac{1}{4}\right)}^2$

${(x+\frac{1}{4})}^2=2+\frac{1}{16}$

${(x+\frac{1}{4})}^2=\frac{33}{16}$

Taking square root on both sides

$\Rightarrow x+\frac{1}{4}=\pm \frac{\sqrt{33}}{4}$

$\Rightarrow x=\pm \frac{\sqrt{33}}{4}-\frac{1}{4},$ $x=\frac{-\sqrt{33}}{4}-\frac{1}{4}$

$\Rightarrow x=\pm \frac{\sqrt{33-1}}{4},$ $x=\frac{-\sqrt{33-1}}{4}$

$c).$ $4x^2+4\sqrt{32}+3=0$

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

$x^2+\sqrt{3}x=\frac{-3}{4}$

Adding $\left(\frac{\sqrt{3}}{2}\right)62$ on both sides

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

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

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

$x=\frac{-\sqrt{3}}{2},\frac{-\sqrt{3}}{2}$

same roots.

$d).$ $2x^2+x+4=0$

$\Rightarrow x^2+\frac{x}{2}+2=0$

$x^2+\frac{x}{2}=-2$

Adding ${\left(\frac{1}{4}\right)}^2$ on both sides

$\Rightarrow x^2+\frac{x}{2}+{\left(\frac{1}{4}\right)}^2=-2+\left(\frac{1}{4}\right)62$

$\Rightarrow {(x+\frac{1}{4})}^2=-2+\frac{1}{16}$

$\Rightarrow {(x+\frac{1}{4})}^2=-\frac{-31}{16}$

Hence, solved.