Question

If $x=\sqrt{(p+q)}+\sqrt{(p-q)}÷\left[\sqrt{(p+q)}-\sqrt{(p-q)}\right]$ then find the value of $q{x}^2-2px+q$

Answer 1

$x=\sqrt{(p+q)}+\sqrt{(p-q)}÷\left[\sqrt{(p+q)}-\sqrt{(p-q)}\right]$

On rationalizing, we get

$x=\frac{\sqrt{(p+q)}+\sqrt{(p-q)}}{\sqrt{(p+q)}-\sqrt{(p-q)}}\times \frac{\sqrt{(p+q)}+\sqrt{(p-q)}}{\sqrt{(p+q)}+\sqrt{(p-q)}}$

Using the identities, $(a+b{)}^2=a^2+b^2+2ab\&(a+b\left)\right(a-b)=(a^2-b^2)$

$x=\frac{[\sqrt{(p+q)}+\sqrt{(p-q)}{]}^2}{[\sqrt{p+q}{]}^2-[\sqrt{p-q}{]}^2}$

$x=\frac{[\sqrt{(p+q)}{]}^2+[\sqrt{(p-q)}{]}^2+[2\times \sqrt{(p+q)}\sqrt{(p-q)}]}{(p+q)-(p-q)}$

$x=\frac{[(p+q)+(p-q)+2\sqrt{(p^2-q^2)}]}{p+q-p+q}$

$x=\frac{[2p+2\sqrt{(p^2-q^2)}]}{2q}$

$x=\frac{2[p+\sqrt{(p^2-q^2)}]}{2q}$

$x=\frac{[p+\sqrt{(p^2-q^2)}]}{q}$

$qx=p+\sqrt{(p^2-q^2)}$

$qx-p=\sqrt{(p^2-q^2)}$

on squaring both sides, we get

$\Rightarrow q^2{x}^2+p^2-2pqx=p^2-q^2$

$\Rightarrow q^2{x}^2-2pqx+q^2=0$

$\Rightarrow q(q{x}^2-2px+q)=0$

$\therefore q{x}^2-2px+q=0$

Hence, the value of $q{x}^2-2px+q=0$