Difference of squares: Using fractions

$(2x)^2-3=6$

First we rearrange the equation and set it to 0

$4x^2-3-6=0$

$4x^2-9=0$

We then apply the shortcut multiplication formula:

$4(x^2-\frac{9}{4})=0$

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

$(x-\frac{3}{2})(x+\frac{3}{2})=0$

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

$±\frac{3}{2}$

$\frac{2x^2-32}{8}=\frac{x+4}{2}$

6

$\frac{(x+7)(x-7)}{3}=-11-x^2$

±2

Solve the following:

$\frac{x(x^2-9)}{x^2+3x}=0$

3

$\frac{x^2-64}{x-8}=\frac{17(x+8)}{64-x^2}$

$±\sqrt{47}$