Question

Its a humble request... please anyone do it.. with steps​

Answer 1

The denominator has radical.

Since the denominator is in the form of a-b, we should multiply a+b to rationalize it.

$\sf{\rightarrow(a+b)(a-b)=a^2-b^2\:[Denominator\:Rationalization]}$

Given fraction:

$\sf{=\dfrac{\sqrt{p+2q}+\sqrt{p-2q}}{\sqrt{p+2q}-\sqrt{p-2q}}\times \dfrac{\sqrt{p+2q}+\sqrt{p-2q}}{\sqrt{p+2q}+\sqrt{p-2q}}}$

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

$\sf{=\dfrac{(p+2q)+2\sqrt{(p+2q)(p-2q)} +(p-2q)}{4q} }$

$\sf{=\dfrac{\cancel{2}p+\cancel{2}\sqrt{(p+2q)(p-2q)} }{4q} }$

$\sf{=\dfrac{p+\sqrt{(p+2q)(p-2q)} }{2q}}$

So the value of x is the following:

• $\sf{x={\dfrac{p+\sqrt{(p+2q)(p-2q)} }{2q} }}$

Now let's simplify the given equation:

→ $\sf{2qx=p+\sqrt{(p+2q)(p-2q)} }$

→ $\sf{2qx-p=\sqrt{(p+2q)(p-2q)} }$

→ $\sf{(2qx-p)^2=(p+2q)(p-2q)}$

→ $\sf{4q^2x^2-4pqx+\cancel{p^2}=\cancel{p^2}-4q^2}$

→ $\sf{\cancel{4}q^2x^2-\cancel{4}pqx+\cancel{4}q^2=0}$

→ $\sf{\cancel{q^2}x^2-p\cancel{q}x+\cancel{q^2}=0}$

$\boxed{\sf{\therefore{qx^2-px+q=0}}}$

Therefore the equality is shown.