Question : If x={√[a+2b] + √[a-2b]} / {√[a+2b] - √[a-2b]}, prove that bx²-ax+b=0
Doubt by Veer
Solution :
Rationalising the denominator
2bx-a=√(a²-4b²)
SBS
(2bx-a)²=a²-4b²
4b²x²+a²-4abx=a²-4b²
4b²x²-4abx=-4b²
4b²x+4b²-4abx=0
4b(bx²+b-ax)=0
4b(bx²-ax+b)=0
4b=0 OR bx²-ax+b=0
b=0/4 OR bx²-ax+b=0
b=0 OR bx²-ax+b=0
but b≠0
Hence bx²-ax+b=0.