Solve the equation : 4x^2 -4a^2x + (a^4 -b^4)=0
Answers
Answered by
28
4x²-4a²x+(a^4 -b^4)=0
⇒4x²-4a²x + (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
therefore
2x-(a²+b²)=0 or 2x-(a²-b²)=0
x= (a²+b²)/2 or x= x= (a²-b²)/2
⇒4x²-4a²x + (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
therefore
2x-(a²+b²)=0 or 2x-(a²-b²)=0
x= (a²+b²)/2 or x= x= (a²-b²)/2
Similar questions