Question

can anybody answer this?

please

Answer 1

x= √a+2b + √ a-2b / √a+2b-√a-2b

prove :-- bx^2- ax + b = 0

take componendo and devedendo rule
(x +1)/( x -1) = {√(a +2b) +√(a -2b)+√(a +2b)-√(a -2b)}/{ √(a +2b) +√(a -2b)-√(a +2b) +√( a -2b)}

(x +1)/(x -1) = √(a +2b)/√(a -2b)

take square both sides

(x +1)²/(x -1)² = (a +2b)/(a -2b)

(a -2b)(x +1)² = (a +2b)(x -1)²

(a -2b)x² +2(a -2b)x +(a -2b) = (a +2b)x²-2(a +2b)x +( a +2b)

( a +2b -a +2b)x² -2x(a +2b+a-2b)x +( a+2b-a +2b)


4bx² -4ax + 4b = 0

bx² -ax + b =0

hence proved

this is your answer.....☺☺☺☺☺