Question

HOLA!!! PLZ SOLVE THIS ONE!!!!!
Solve this by quadratic equation:
4x² - 4ax + (a²- b²)​=0
I want a detailed explanation..


the answer which would satisfy me would be brainliest......​

Answer 1

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

Hope this helps you!

Answer 2

$\huge{\underline{\underline{\mathtt{Answer}}}}$

Given Equation,

4x² - 4ax + (a² - b²) = 0

→4x² - 4ax + (a+b)(a-b) = 0

Here,the middle term is -4a

We need to express the constant term in terms of the middle term

i.e., -4a = [(a+b)+(a-b)]

•We obtain the equation,

4x² -2[(a+b) + (a-b)]x + (a+b)(a-b)=0

•By splitting the middle term,

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

→2x[2x - (a+b)] -(a-b)[2x - (a+b)]=0

→[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 (a-b)/2

• Thus,the roots are (a+b)/2 and (a-b)/2