Question

if x = root 7 + root 6 / root 7 - root 6 , then find value of (x+1/x)2

Answer 1

x =(√7+√6)/(√7-√6)

x + 1/x = (√7+√6)/(√7-√6)+(√7-√6)/(√7+√6)
= {(√7+√6)² + (√7-√6)²}/{(√7+√6)(√7-√6)}
= 2(7+6)/{7-6}
= 2× 13/1
= 26

(x + 1/x)² = 26² = 676

I hope you understand the approach.
All the best!

Answer 2

The value of $(x+\frac{1}{x})^{2}$ is $676$

Step-by-step explanation:

Given,

$x=\frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}-\sqrt{6}}$ then find value of $(x+\frac{1}{x})^{2}$

∴ $x=\frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}-\sqrt{6}}$

⇒$x=\frac{(\sqrt{7}+\sqrt{6})(\sqrt{7}+\sqrt{6})}{(\sqrt{7}-\sqrt{6})(\sqrt{7}+\sqrt{6})}$

⇒$x=\frac{(\sqrt{7}+\sqrt{6})^{2}}{(\sqrt{7})^{2} -(\sqrt{6})^{2} }$

⇒$x=(\sqrt{7})^{2}+(\sqrt{6})^{2}+2\sqrt{42}$

⇒$x=13+2\sqrt{42}$

Similarly,

$\frac{1}{x}=\frac{1}{\frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}-\sqrt{6}}}$

⇒$\frac{1}{x}=\frac{\sqrt{7}-\sqrt{6}}{\sqrt{7}+\sqrt{6}}$

⇒$\frac{1}{x}=13-2\sqrt{42}$

∴ $(x+\frac{1}{x})^{2}$

$=(13+2\sqrt{42}+13-2\sqrt{42})^{2}$

$=(13+13)^{2}$

$=(26)^{2}$

$=676$

So, The value of $(x+\frac{1}{x})^{2}$ is $676$