Question

Solve the following quadratic equation for $x$:
$4x^2+4bx-(a^2-b^2)=0$

Answer 1

We know that for any quadratic equation of the form $a{x}^2+bx+c=0$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Given equation is

$4x^2+4bx-a^2+b^2=0$

Hence, substituting the values of a,b and c in the above equation we get,

$x=\frac{-4b\pm \sqrt{(4b{)}^2-4\times 4\times (-a^2+b^2)}}{2\times 4}$

$=\frac{-4b\pm \sqrt{16b^2+16a^2-16b^2}}{8}$

$=\frac{-4b\pm 4a}{8}$

$=\frac{-b\pm a}{2}$

So, $x=\frac{-b+a}{2}$

or

$x=\frac{-b-a}{2}$

$=-\frac{b+a}{2}$

Hence, these are the values of x.