Question

x=rootm+n+rootm-n/rootm+n-rootm-n find nx^2-2mx+n
please help me fast
why even rational doesnt know this answer..

Answer 1

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

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}$

Answer 2

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.