Question

Solve each of the following quadratic equations:
$3\sqrt{7}{x}^2+4x-\sqrt{7}=0$

Answer 1

$3\sqrt{7}{x}^2+4x-\sqrt{7}=0$

$3\sqrt{7}{x}^2+(7-3)x-\sqrt{7}=0$

$3\sqrt{7}{x}^2+7x-3x-\sqrt{7}=0$

$\sqrt{7}x(3x+\sqrt{7})-1(3x+\sqrt{7})=0$

$(3x+\sqrt{7})(\sqrt{7}x-1)=0$
when
$3x+\sqrt{7}=0$
$3x=-\sqrt{7}$

$\Rightarrow x=\frac{-\sqrt{7}}{3}$

Or

$\sqrt{7}\times x-1=0$

$\Rightarrow$ $\sqrt{7}\times x=1$
$\Rightarrow$ $x=\frac{1}{\sqrt{7}}$

$\Rightarrow eitherx=\frac{1}{\sqrt{7}}$
or x= $\frac{-\sqrt{7}}{3}$