Question

If $x=\frac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}-\sqrt{a-2b}}$ , then show that $b{x}^2-ax+b=0$.

Answer 1

$$\begin{gathered} \text{Given}, \\ x=\frac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}-\sqrt{a-2b}} \\ \mathrm{Rationalising}\mathrm{the}\mathrm{denominator},\mathrm{we}\mathrm{get}: \\ x=\frac{\left(\sqrt{a+2b}+\sqrt{a-2b}\right)}{\left(\sqrt{a+2b}-\sqrt{a-2b}\right)}\times \frac{\left(\sqrt{a+2b}+\sqrt{a-2b}\right)}{\left(\sqrt{a+2b}+\sqrt{a-2b}\right)} \\ \Rightarrow x=\frac{{\left(\sqrt{a+2b}+\sqrt{a-2b}\right)}^2}{{\left(\sqrt{a+2b}\right)}^2-{\left(\sqrt{a-2b}\right)}^2} \\ \Rightarrow x=\frac{{\left(\sqrt{a+2b}\right)}^2+{\left(\sqrt{a-2b}\right)}^2+2\sqrt{a+2b}\sqrt{a-2b}}{\left(a+2b\right)-\left(a-2b\right)} \\ \Rightarrow x=\frac{a+2b+a-2b+2\sqrt{a^2-{\left(2b\right)}^2}}{4b} \\ \Rightarrow x=\frac{2a+2\sqrt{a^2-4b^2}}{4b} \\ \Rightarrow x=\frac{a+\sqrt{a^2-4b^2}}{2b} \\ \Rightarrow 2bx=a+\sqrt{a^2-4b^2} \\ \Rightarrow 2bx-a=\sqrt{a^2-4b^2} \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides}\mathrm{we}\mathrm{get}: \\ {\left(2bx-a\right)}^2={\left(\sqrt{a^2-4b^2}\right)}^2 \\ \Rightarrow 4b^2{x}^2+a^2-4abx=a^2-4b^2 \\ \Rightarrow 4b^2{x}^2-4abx+4b^2=0 \\ \Rightarrow 4b\left(b{x}^2-ax+b\right)=0 \\ \mathrm{Therefore},b{x}^2-ax+b=0 \\ \mathrm{Hence},\mathrm{proved}. \end{gathered}$$