Question

if x=root 3- root2 then find the value of [x-1/x]^2

Answer 1

GIVEN:

★$x=\sqrt{3}-\sqrt{2}$

TO FIND:

★The value of $(x-\dfrac{1}{x}) ^{2}$

CONCEPT USED:

★We would first find $\dfrac{1}{x}$ and then we will subtract $\dfrac{1}{x}$ from $x$ and then square the number obtained.

ANSWER:

We have,

=>$x=\sqrt{3}-\sqrt{2}$

=>$\dfrac{1}{x}=\dfrac{1}{\sqrt{3}-\sqrt{2}}$

On rationalising the denominator,

=>$\dfrac{1}{x}=\dfrac{1(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}$

=>$\dfrac{1}{x}=\dfrac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3})^{2}-(\sqrt{2}^{2})}$

$\large\green{\boxed{(a+b)(a-b)=a^{2}-b^{2}}}$

=>$\dfrac{1}{x}=\dfrac{\sqrt{3}+\sqrt{2}}{3-2}$

=>$\dfrac{1}{x}=\dfrac{\sqrt{3}+\sqrt{2}}{1}$

=>$\dfrac{1}{x}=\sqrt{3}+\sqrt{2}$

$\large\purple{\boxed{\dfrac{1}{x}=\sqrt{3}+\sqrt{2}}}$

______________________________________

Hence,

=$(x-\dfrac{1}{x}) ^{2}$

=$[\sqrt{3}-\sqrt{2}-(\sqrt{3}+\sqrt{2}) ^{2}]$

=$[\sqrt{3}-\sqrt{2}-\sqrt{3}-\sqrt{2})]^{2}$

=$(-2\sqrt{2}) ^{2}$

=$8$

$\huge\orange{\boxed{.°.(x-\dfrac{1}{x}) ^{2}=8}}$