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2015-04-25T12:12:06+05:30

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x=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}=\frac{(\sqrt{m+n}+\sqrt{m-n})^2}{\sqrt{m+n}^2-\sqrt{m-n}^2}=\frac{m+\sqrt{m^2-n^2}}{n}

similarly \frac{1}{x}=\frac{m-\sqrt{m^2-n^2}}{n}

Now consider the given expression
nx^2-2mx+n
Factor out x and get
x\left(nx-2m+n\cdot\frac{1}{x}\right)
=x\left(n\cdot\frac{m+\sqrt{m^2-n^2}}{n}-2m+n\cdot\frac{m-\sqrt{m^2-n^2}}{n}\right)=x\left(m+\sqrt{m^2-n^2}-2m+m-\sqrt{m^2-n^2}\right)\\=x\left(0\right)\\=\boxed{0}
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2015-04-26T01:43:24+05:30

This Is a Certified Answer

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Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Rationalize the given expression in m and n by multiplying the denominator and number ator with the conjugate of the denominator.

x=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}\\\\=\frac{(\sqrt{m+n}+\sqrt{m-n})(\sqrt{m+n}+\sqrt{m-n})}{(\sqrt{m+n}-\sqrt{m-n})(\sqrt{m+n}+\sqrt{m-n})}\\\\=\frac{(\sqrt{m+n})^2+(\sqrt{m-n})^2+2\sqrt{(m+n)(m-n)}}{(\sqrt{m+n})^2-(\sqrt{m-n})^2}\\\\=\frac{m+n+m-n+2\sqrt{m^2-n^2}}{m+n-(m-n)}\\\\=\frac{2\ m\ +\ 2\ \sqrt{m^2-n^2}}{2\ n},\ \ \ ---(1)


Let\ P(x)=nx^2-2mx+n\\\\Roots\ of\ quadratic\ equation\ nx^2-2mx+n=0,\ \ \ are:\\.\ \ \ x=\frac{2m+-\sqrt{(2m)^2-4*n*n}}{2n}=\frac{2m+-2\sqrt{m^2-n^2}}{2n},\ \ ---(2)

Comparing the value of x we have in equation (1) and the root of the quadratic expression in (2), we see that they are same.  Hence, the value of quadratic expression is zero for that value of x.

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