Question

Solve the following equations by factorization method:
4 x^2 - 4 a^2 x + ( a^4 - b^4 ) = 0

Answer 1

Solution-

Here, we have

4x² - 4a²x + (a⁴ - b⁴) = 0

Constant term = a⁴ - b⁴ = (a² - b²) (a² + b²)

Coefficient of middle term = - 4a²

Coefficient of middle term = - 4a² = - [2(a² + b²) + 2(a² - b²)]

Then,

⇒ 4x² - 4a²x + (a⁴ - b⁴) = 0

⇒ 4x² - [2(a² + b²) + 2(a² - b²)]x + (a² - b²) (a² + b²) = 0

⇒ 4x² - 2(a² + b)x - 2(a² - b²)x + (a² - b²) (a² + b²) = 0

⇒ [4x² - 2(a² + b²)x] - [2(a² - b²)x - (a² - b²) (a² + b²) = 0

⇒ 2x[2x - (a² + b²)] - (a² - b²) [2x - (a² + b²)] = 0

⇒ [2x - (a² + b²)] [2x - (a² - b²)] = 0

⇒ 2x - (a² + b²) = 0 or 2x - (a² - b²) = 0

⇒ x = a² + b²/2, a² - b²/2

⇒ x = a² + b²/2, a² - b²/2

Hence, x = a² + b²/2, a² - b²/2.

Answer 2

Answer:

me Kuch bhi answer Karu tuze kya Karna he correct du ya bakwas

Tu aapna dekhe pehele ....hehe