Question

if x = ${\displaystyle \frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}}$ using properties of proportion show that ${\displaystyle x^2-2ax+1=0}$

Answer 1

Consider ${\displaystyle x=\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}}$

${\displaystyle \Rightarrow \frac{x}{1}=\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}}$

By using componando and dividendo, we have

${\displaystyle \Rightarrow \frac{x+1}{x-1}=\frac{(\sqrt{a+1}+\sqrt{a-1})+(\sqrt{a+1}-\left(\sqrt{a-1}\right))}{(\sqrt{a+1}+\sqrt{a-1})-(\sqrt{a+1}-\sqrt{a-1})}}$

${\displaystyle \Rightarrow \frac{x+1}{x-1}=\frac{2\sqrt{a+1}}{2\sqrt{a-1}}=\frac{\sqrt{a+1}}{\sqrt{a-1}}}$

Squaring both sides we get

${\displaystyle \Rightarrow \frac{{(x+1)}^2}{{(x-1)}^2}=\frac{{\left(\sqrt{a+1}\right)}^2}{{\left(\sqrt{a-1}\right)}^2}}$

${\displaystyle \Rightarrow \frac{x^2+2x+1}{x^2-2x+1}=\frac{a+1}{a-1}}$

Again using componando and dividendo, we get

${\displaystyle \Rightarrow \frac{(x^2+2x+1)+(x^2-2x+1)}{(x^2+2x+1)-(x^2-2x+1)}=\frac{(a+1)+(a-1)}{(a+1)-(a-1)}}$

${\displaystyle \Rightarrow \frac{2x^2+2}{4x}=\frac{2a}{2}\Rightarrow \frac{x^2+1}{2x}=\frac{a}{1}}$

${\displaystyle \Rightarrow x^2+1=2ax}$

${\displaystyle \Rightarrow x^2-2ax+1=0}$

Answer 2

${\displaystyle x=\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}}$

⇒ ${\displaystyle \frac{x}{1}=\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}}$

By componendo and dividendo

${\displaystyle \frac{x+1}{x-1}=\frac{\sqrt{a+1}+\sqrt{a-1}+\sqrt{a+1}-\sqrt{a-1}}{\sqrt{a+1}+\sqrt{a-1}-\sqrt{a+1}-\sqrt{a-1}}}$

⇒ ${\displaystyle \frac{x+1}{x-1}= \frac{\sqrt{a+1}}{\sqrt{a-1}}}$

⇒ ${\displaystyle \frac{{(x+1)}^2}{{(x-1)}^2}=\frac{\sqrt{a+1}}{\sqrt{a-1}}}$ ...(by duplicate ratio)

⇒ ${\displaystyle \frac{x^2+2x+1}{x^2-2x+1}=\frac{a+1}{a-1}}$

Again by componendo and dividendo

${\displaystyle \Rightarrow \frac{x^2+2x+1+x^2-2x+1}{x^2+2x+1-x^2+2x-1}}$

= ${\displaystyle \frac{a+1+a-1}{a+1-a-1}}$

⇒ ${\displaystyle \frac{x^2+1}{2x}=\frac{a}{1}}$

⇒ ${\displaystyle x^2+1=2ax}$

⇒ ${\displaystyle x^2-2ax+1=0}$
Hence proved.