If x1=√a+3b+√a−3b√a+3b−√a−3b then, which one of the following statements is true?
3bx2+3b−2ax=0
x1=√a+3b+√a−3b√a+3b−√a−3b
Applying componendo and dividendo
x+1x−1=√a+3b+√a−3b+√a+3b−√a−3b√a+3b+√a−3b−√a+3b+√a−3b
= 2√a+3b2√a−3b
= √a+3b√a−3b
Squaring on both sides, we get:
(x+1)2(x−1)2=(√a+3b)2(√a−3b)2
= (x+1)2(x−1)2=a+3ba−3b
Again, applying componendo and dividendo
(x+1)2+(x−1)2(x+1)2−(x−1)2=a+3b+a−3ba+3b−a+3b
2(x2+1)4x=2a6b
=(x2+1)2x=a3b
=3bx2+3b=2ax
= 3bx2+3b−2ax=0