Question

Let's see who can solve it:
Best answer will be marked as brainliest......
solve the value of x:
4x^2-4a^2+(a^4-b^4)=0

(with solution)

Answer 1

Given Equation is 4x^2 - 4a^2 + (a^4 - b^4) = 0

4x^2 - 4a^2 + a^4 - b^4 = 0

a^4 - b^4 + 4x^2 - 4a^2 = 0

a^4 - b^4 + 4x^2 = 4a^2

-b^4 + 4x^2 = -a^4 + 4a^2

4x^2 = -a^4 + 4a^2 + b^4

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

$x = \sqrt{ \frac{-a^4 + 4a^2 + b^4}{4} } , x = -\sqrt{ \frac{-a^4+4a^2+b^4}{4} }$

$x = \frac{ \sqrt{-a^4 + 4a^2 + b^2} }{2} , x = - \frac{ \sqrt{-a^4 + 4a^2 + b^2} }{2}$



Hope this helps!

Answer 2

4x² - 4a² + a⁴ - b⁴ = 0
4x² = 4a² - a⁴ + b⁴
x² = (4a² - a⁴ + b⁴)/4
take square root both sides,
x = ±√(4a² - a⁴ + b⁴)/√4
x = ±√(4a² - a⁴ + b⁴)/2

hence, √(4a² - a⁴ + b⁴)/2 and -√(4a² - a⁴ + b⁴)/2 are the values of x