Question

☆ Solve for a comparing with the
Standard equation in
→ 4x² - 4 ax +(a²-b^2)=0​

Answer 1

Answer:

Given: 4x² - 4ax + (a² - b²) = 0.

Here in this equation, constant term = (a² - b²) = (a+b)(a-b)

Coefficient of middle term= - 4a

Also, Coefficient of the middle term= -[2(a+b)+2(a+b)]

4x² - 4ax + (a² - b²) = 0.

4x² -[2(a+b)+2(a-b)]x + (a+b)(a-b)= 0

4x² - 2(a+b)x - 2(a-b)x + (a+b)(a-b)= 0

[4x² - 2(a+b)x ] - [ 2(a- b)x + (a+b)(a-b)]= 0

2x [ 2x-(a+b)] -(a-b)[2x -  (a+b)]

[2x -  (a+b)] [2x-(a-b)]= 0

[2x -  (a+b)] = 0  or  [2x-(a-b)]= 0

2x = a + b   or    2x = a-b

x =( a+b)/2 or  x= (a-b)/2