Question

4x^2-4ax+(a^2-b^2)=0 solve the quadratic equation step by step with explanation ​

Answer 1

Question -

Solve for x-

4x²-4ax+(a²-b²)=0

Solution -

The given quadratic equation is 4x²-4ax+(a²-b²)=0

=> 4x²-4ax+(a²-b²)=0

=> 4x²-4ax+(a-b)(a+b)=0

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

=> 4x²+(2b-2a)x-(2a+2b)x+(a-b)(a+b)=0

=> 4x²+2(b-a)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

Now,

$2x - (a - b) = 0$

$2x = a - b$

$x = \frac{a - b}{2}$

And,

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

$2x = a + b$

$x = \frac{a + b}{2}$

Therefore the value of x is (a-b)/2 and (a+b) /2

Answer 2

Answer:

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