Question

Solve the quadratic equation for x:
4x²- 4a²x + ($a^{4}$-$b^{4}$) = 0

Answer 1

4x²- 4a²x + ($a^{4}$-$b^{4}$)

Here,It the case of above quadratic equation.

a. =4

b = - 4a²

c. = ($a^{4}$-$b^{4}$)

We know that,

D. = b² - 4 ac
=>D = (4a²)² -4(4)($a^{4}$-$b^{4}$)
D =
$d \: \: = 16 {a}^{4} \: - 16 {a}^{4} \: \: - 16 {b}^{4}$

$d = 16 {b}^{4}$

Root of above equation
can be
$\binom{ - b - \sqrt{d} }{2c}$

OR

$\binom{ - b + \sqrt{d} }{2a}$
$= > x = \binom{1 - {b}^{2} }{2} \: \: or \: \: \binom{1 + {b}^{2} }{2}$