Then what is bx2 2ax b equal to b0 ........

Quantitative Aptitude (CDS) Quadratic Equations

$If\; x=\;\frac{\sqrt{a+b}-\sqrt{a-b}}{\sqrt{a+b}+\sqrt{a-b}}$

Then what is bx²-2ax+b equal to (b $\neq$ 0)?

2ab

$As,\: x= \frac{\sqrt{a+b}-\sqrt{ab}}{\sqrt{a+b}-\sqrt{ab}}$

Rationalizing

$x= \frac{\sqrt{a+b}-\sqrt{ab}}{\sqrt{a+b}-\sqrt{a+b}}\times \frac{\sqrt{a+b}-\sqrt{a+b}}{\sqrt{a-b}-\sqrt{a+b}}$

$x= \frac{a}{b}-\frac{\sqrt{a^{2}-b^{2}}}{b}$

$bx^{2}= b[ \frac{a}{b}-\frac{\sqrt{a^{2}-b^{2}}}{b}]^{2}$

$=-\frac{2a^{2}}{b}+\frac{2a\sqrt{a^{2}-b^{2}}}{b}$

Now,

$bx^{2}-2ax+b = \frac{a^{2}}{b}+\frac{a^{2}- b^{2}}{b}-\frac{2a\sqrt{a^{2}-b^{2}}}{b}-\frac{2a^{2}}{b}+\frac{2a\sqrt{a^{2}-b^{2}}}{b}+b$

= 0

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