Question

Solve the following quadratic equation in x :
a^2b^2x^2 - 4b^4x + 3a^4x - 12a^2b^2 = 0

Answer 1

a²b²x²-(4b⁴-3a⁴)x-12a²b²=0

=>x={(4b⁴-3a⁴)±√((4b⁴-3a⁴)²+48a⁴b⁴)}/2a²b²

=>x={(4b⁴-3a⁴)±√((2b²-a²√3)²(a²√3+2b²)²+48a⁴b⁴)}/2a²b²

Answer 2

$a^2b^2x^2 - 4b^4x + 3a^4x - 12a^2b^2 = 0 \\ \\ \Rightarrow\ (ab)^2x^2-(4b^4-3a^4)x-12(ab)^2=0$

$Discriminant \\ \\ = (-(4b^4-3a^4))^2-(4 \cdot (ab)^2 \cdot (-12(ab)^2)) \\ \\ = (16b^8-24(ab)^4+9a^8)-(4(ab)^2 \cdot (-12(ab)^2)) \\ \\ = (16b^8-24(ab)^4+9a^8)-(-48(ab)^4) \\ \\ = 16b^8-24(ab)^4+9a^8+48(ab)^4 \\ \\ = 16b^8+24(ab)^4+9a^8 \\ \\ = (4b^4+3a^4)^2$

$x=\frac{-(-(4b^4-3a^4))\pm\sqrt{(4b^4+3a^4)^2}}{2(ab)^2} \\ \\ x=\frac{4b^4-3a^4\pm(4b^4+3a^4)}{2a^2b^2} \\ \\ x=\frac{4b^4-3a^4+(4b^4+3a^4)}{2a^2b^2} \ \ \ \ \ ; \ \ \ \ \ x=\frac{4b^4-3a^4-(4b^4+3a^4)}{2a^2b^2} \\ \\ x=\frac{4b^4-3a^4+4b^4+3a^4}{2a^2b^2} \ \ \ \ \ \ \ ; \ \ \ \ \ x=\frac{4b^4-3a^4-4b^4-3a^4}{2a^2b^2} \\ \\ x=\frac{8b^4}{2a^2b^2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ; \ \ \ \ \ x=\frac{-6a^4}{2a^2b^2}$

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

$Hope this helps. \\ \\ \\ Thank you. Have a nice day. :-)$