Math, asked by amoli2004, 10 months ago

Find the roots 4x^2-4a^2x+a^4-b^4 =0​

Answers

Answered by Anonymous
12

Answer:

4x²-4a²x+(a⁴-b⁴)=0

=> 4x²-2{(a²+b²)+(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

Either, 2x-(a²+b²)=0

2x = a²+b²

x = (a²+b²)/2.

OR 2x-(a²-b²)=0

2x = a²-b²

x = (a²-b²)/2.

x = (a²+b²)/2 , (a²-b²)/2

Hope it will help you.


amoli2004: thanks
Anonymous: Your welcome.
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