can anybody answer this
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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
this is your answer.......☺☺☺☺
hence proved
(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
this is your answer.......☺☺☺☺
hence proved
futsci:
thanx very much. may god bless you
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