Question

using quadratic formula solve for the following
a^2b^2x^2-(4b^4-3a^4)x -12a^2b^2=0
​

Answer 1

Answer:

${a}^{2} {b}^{2} \times x {}^{2} -(4b{}^{4} - 3a {}^{4} )x - 12 {a}^{2} {b}^{2} = 0 \\ = > x = \frac{4 {b}^{4} - 3 {a}^{4} \pm \sqrt{(4 {b}^{4} - 3 {a}^{4} ) {}^{2} - 4 \times {a}^{2} {b}^{2} \times - 12 {a}^{2}b {}^{2} } }{2 \times {a}^{2} {b}^{2} } \\ = > x = \frac{4 {b}^{4} - 3 {a}^{4} \pm \sqrt{16 {b}^{8} - 24 {a}^{4} {b}^{4} + 9 {a}^{8} + 48 {a}^{4} {b}^{4} } }{2 {a}^{2} {b}^{2} } \\ = > x = \frac{4b {}^{4} - 3a {}^{4} \pm \sqrt{9 {a}^{8} + 24a {}^{4} {b}^{4} + 16 {b}^{8} } }{2 {a}^{2} {b}^{2} } \\ = > x = \frac{4 {b}^{4} - 3 {a}^{4} \pm \sqrt{(3 {a}^{4} ) {}^{2} + 2 \times 3a {}^{4} \times 4 {b}^{4} + (4 {b}^{4}) {}^{2} } }{2 {a}^{2} {b}^{2} } \\ = > x = \frac{4 {b}^{4} - 3 {a}^{4} \pm \ \sqrt{(3 {a}^{4} + 4b {}^{4}) {}^{ {}^{2} } } }{2 {a}^{2}b {}^{2} } \\ = > x = \frac{4 {b}^{4} - 3 {a}^{4} \pm(3 {a}^{4} + 4 {b}^{4} )}{2 {a}^{2} {b}^{2} }$

either,

$x = \frac{4 {b}^{4} - 3 {a}^{4} + 3 {a}^{4} + 4 {b}^{4} }{2 {a}^{2} {b}^{2} } \\ = > x = \frac{8 {b}^{4} }{2 {a}^{2}b {}^{2} } \\ = > x = \frac{4 {b}^{2} }{ {a}^{2} }$

or,,,,

$x = \frac{4 {b}^{4} - 3 {a}^{4} - 3 {a}^{4} - 4b {}^{4} }{2 {a}^{2} {b}^{2} } \\ = > x = \frac{ - 6 {a}^{4} }{2 {a}^{2} {b}^{2} } \\ = > x = \frac{ - 3{a}^{2} }{ {b}^{2} }$

so,

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