Find the roots 4x^2-4a^2x+a^4-b^4 =0
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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
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