Question

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:
$a^2{b}^2{x}^2-(4b^4-3a^4)x-12a^2{b}^2=0,a\ne 0andb\ne 0$

Answer 1

$a^2{b}^2{x}^2-(4b^4-3a^4)x-12a^2{b}^2=0$

Here $A=a^2{b}^2$
$B=-(4b^4-3a^4)$
$C=-12a^2{b}^2$

$D=discriminant=B^2-4AC$
$=\left[\right(-(4b^4-3a^4{)}^2-4\times [a^2{b}^2]\times [-12a^2{b}^2]$
$=[16b^8+9a^8-24a^4{b}^4]+48a^4{b}^4$
$=16b^8+9a^8-24a^4{b}^4$
$=(4b^4{)}^2+2\times 4b\times 3a^2+(3a^2{)}^2$
$=(4b^4+3a^4{)}^2[\because (a+b{)}^2=a^2+b^2+2ab]$

Now, let us use the quadratic formula

$x=\frac{-B\pm \sqrt{B^2-4AC}}{2A}$

$x=\frac{-B\pm \sqrt{D}}{2A}$

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

$x=\frac{(4b^4-3a^4)\pm 4b^4+3a^4}{2a^2{b}^2}$

now taking positive sign,

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

$x=\frac{8b^4}{2a^2{b}^2}$

$x=\frac{4b^2}{a^2}$

taking negative value :

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

$x=\frac{-6a^4}{2a^2{b}^2}$

$x=\frac{-3a^2}{b^2}$

$\therefore X=\frac{4b^2}{a^2},\frac{-3a^2}{b^2}$