If `x=(sqrt(a+2b)+sqrt(a-2b))/(sqrt(a+2b)-sqrt(a-2b))`, then prove that `b ^2-ax+b=0`

Explanation

We have, `x=(sqrt(a+2b)+sqrt(a-2b))/(sqrt(a+2b)-sqrta-2b)=(sqrt(a+2b)+sqrt(a-2b))/(sqrt(a+2b)-sqrta-2b)xx(sqrt(a+2b)+sqrt(a-2b))/(sqrt(a+2b)-sqrta-2b)`
`=((sqrt(a+2b)+sqrt(a-2b))^(2))/((sqrt(a+2b)-sqrta-2b)^(2))=(a+2b+a-2b+2sqrt((a+2b)(a-2b)))/((a+2b)-(a-2b))`
`(2a+2sqrt(a^(2)-4b^(2)))/(4b)rArr" "x=(a+sqrt(a^(2)-4b^(2)))/(2b)`
`rArr" "2bx=a+sqrt(a^(2)-4b^(2))" "rArr2bx-a=sqrt(a^(2)-4b^(2))`
Squaring both sides, we get
`=(2bx-a)^(2)=a^(2)-4b^(2)`
`=4b^(2)x^(2)+a^(2)-4abx=a^(2)-4b^(2)`
`=4b^(2)x^(2)-4abx+4b^(2)=0`
`rArr" "4b(bx^(2)-ax+b)=0`
`:." either "b=0orbx^(2)-ax+b=0`
But b cannot be zero as it will not give the real value of x.
`:." "bx^(2)-ax+b=0`