Question

solve by factorization- 4x 2 -4a 2 x + (a 4 -b 4 ) = 0

Answer 1

4x²-4a²x+(a⁴-b⁴)=0
or, 4x²-2{(a²+b²)+(a²-b²)}x+(a²+b²)(a²-b²)=0
or, 4x²-2(a²+b²)x-2(a²-b²)x+(a²+b²)(a²-b²)=0
or, 2x{2x-(a²+b²)}-(a²-b²){2x-(a²+b²)}=0
or, {2x-(a²+b²)}{2x-(a²-b²)}=0
Either, 2x-(a²+b²)=0
or, 2x=a²+b²
or, x=(a²+b²)/2
Or, 2x-(a²-b²)=0
or, 2x=a²-b²
or, x=(a²-b²)/2
∴, x=(a²+b²)/2, (a²-b²)/2 Ans.

Answer 2

Answer:

Step-by-step explanation:

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

Hence, x = a² + b²/2, a² - b²/2.