Question

using the quadratic formula, solve the equation a square b square x square - (4b power 4-3apower4)x-12a square b square =0

Answer 1

$\textbf{Given equation is}$

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

$\text{Then, the quadratic formula is}$

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

$x=\displaystyle\frac{(4b^4-3a^4)\pm\sqrt{(4b^4-3a^4)^2+4(a^2b^2)(12a^2b^2)}}{2a^2b^2}$

$x=\displaystyle\frac{(4b^4-3a^4)\pm\sqrt{(4b^4-3a^4)^2+48a^4b^4}}{2a^2b^2}$

$x=\displaystyle\frac{(4b^4-3a^4)\pm\sqrt{16b^8+9a^8-24a^2b^2+48a^4b^4}}{2a^2b^2}$

$x=\displaystyle\frac{(4b^4-3a^4)\pm\sqrt{16b^8+9a^8+24a^2b^2}}{2a^2b^2}$

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

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

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

$x=\displaystyle\frac{8b^4}{2a^2b^2},\;\displaystyle\frac{-6a^4}{2a^2b^2}$

$\implies\bf\,x=\displaystyle\frac{4b^2}{a^2},\;\displaystyle\frac{-3a^2}{b^2}$

$\therefore\textbf{The solution set is \{ \frac{4b^2}{a^2},\;\frac{-3a^2}{b^2} \}}$

Find more:

Using the quadratic formula to solve 5x = 6x2 – 3, what are the values of x

https://brainly.in/question/14911682

Answer 2

Step-by-step explanation:

1. here is your solution.