Question

plz solve 21 question fast​

Answer 1

I hope it will be answer of your question

Answer 2

Answer:

Step-by-step explanation:

x² -2ax + a² - b² = 0

=> x² -ax - ax  + (a+b)(a-b) = 0

=> x² -ax + bx  -ax - bx + (a+b)(a-b) = 0

=> x² -ax + bx  -x(a+ b)  + (a+b)(a-b) = 0

=> x(x - a + b) - (a+b)(x - (a-b)) = 0

=> (x-a+b) (x - a - b) = 0

x² -4ax + 4a² - b² = 0

=> x² -2ax - 2ax  + (2a)² - b² = 0

=> x² -2ax + bx  -2ax - bx + (2a+b)(2a-b) = 0

=> x² -2ax + bx  -x(2a+ b)  + (2a+b)(2a-b) = 0

=> x(x - 2a + b) - (2a+b)(x - (2a-b)) = 0

=> (x-2a+b) (x - 2a - b) = 0

4x² -4ax + a² - b² = 0

=> (2x)² -2ax - 2ax  + (a+b)(a-b) = 0

=> (2x)² -2ax + 2bx  -2ax - 2bx + (a+b)(a-b) = 0

=> (2x)² -2ax + 2bx  -2x(a+ b)  + (a+b)(a-b) = 0

=> 2x(x - a + b) - (a+b)(2x - (a-b)) = 0

=> (2x-a+b) (2x - a - b) = 0

4x² - 4a²x + a⁴ - b⁴ = 0

=> (2x)² - 2a²x -2a²x + (a² + b²)(a²-b²) = 0

=> (2x)² - 2a²x +2b²x -2a²x - 2b²x + (a² + b²)(a²-b²) = 0

=>  (2x)² - 2a²x +2b²x - 2x(a² + b²) + (a² + b²)(a²-b²) = 0

=> 2x(2x - a² +b²) - (a² + b²)(2x - a²+ b²) = 0

=> (2x - a² - b²)(2x - a²+ b²) = 0