Question

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

Answer 1

The given equation is:
a²b²x² − (4b⁴ − 3a⁴)x − 12a²b² = 0
Comparing it with Ax² + Bx + C = 0, we get:
A = a²b² , B = − (4b⁴ − 3a⁴) and C = − 12a²b²
∴ D = (B² − 4AC)
= (3a⁴ − 4b⁴ )² − 4(a²b²)( − 12a²b² )
= 9a⁸ + 16b⁸ − 24a⁴b⁴ + 48a⁴b⁴
= (3a⁴)² + (4b⁴)² + 2 × 3a⁴ × 4b⁴
= (3a⁴ + 4b⁴)² ≥ 0
Hence, the given equation has real roots. These are
$\alpha =\frac{-B+\sqrt{D}}{2A}=\frac{\left(4b^4-3a^4\right)+\left(3a^4+4b^4\right)}{2a^2{b}^2}=\frac{8b^4}{2a^2{b}^2}=\frac{4b^2}{a^2}$
$\beta =\frac{-B-\sqrt{D}}{2A}=\frac{\left(4b^4-3a^4\right)-\left(3a^4+4b^4\right)}{2a^2{b}^2}=\frac{-6a^4}{2a^2{b}^2}=\frac{-3a^2}{b^2}$
Hence, $\frac{4b^2}{a^2}\mathrm{and}\frac{-3{\mathrm{a}}^2}{{\mathrm{b}}^2}$ are the roots of the given equation.