Question

if x is equals to root 3 minus root 2 find the value of x square + 1 by x square​

Answer 1

x = root3 - root 2
Then
1/x = root 3 + root 2

Now x + 1/x = root 3 - root 2 + root 3 + root 2

X + 1/x = 2 root 3
Square both side

Xsquare + 1 x sauare + 2 = 12
So
x square + 1/x square = 12-2 = 10
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Answer 2

The value of the expression $x^2+\frac{1}{x^2}$ is 10.

Given,

The value of $x=\sqrt{3}-\sqrt{2}$ .

To Find,

The value of the expression: $x^2+\frac{1}{x^2}$.

Solution,

The method of finding the value of the given expression is as follows -

First, we will compute the value of $x^2$.

$x^2=(\sqrt{3}-\sqrt{2})^2=(\sqrt{3})^2+(\sqrt{2} )^2-2*\sqrt{2}*\sqrt{3} \\ =3+2-2\sqrt{2}\sqrt{3} =5-2\sqrt{2}\sqrt{3}$

So, $x^2+\frac{1}{x^2}=5-2\sqrt{2}\sqrt{3}+\frac{1}{5-2\sqrt{2}\sqrt{3}}$

$=5-2\sqrt{2}\sqrt{3}+\frac{5+2\sqrt{2}\sqrt{3}}{(5-2\sqrt{2}\sqrt{3})(5+2\sqrt{2}\sqrt{3})}$ [rationalizing the denominator]

$=5-2\sqrt{2}\sqrt{3}+\frac{5+2\sqrt{2}\sqrt{3}}{5^2-(2\sqrt{2} \sqrt{3} )^2}= 5-2\sqrt{2}\sqrt{3}+\frac{5+2\sqrt{2}\sqrt{3}}{25-24}$ [Since $(a+b)(a-b)=a^2-b^2$]

$=5-2\sqrt{2}\sqrt{3}+\frac{5+2\sqrt{2}\sqrt{3}}{1}=5-2\sqrt{2}\sqrt{3}+5+2\sqrt{2}\sqrt{3}$

$=10$

Hence, the value of the expression $x^2+\frac{1}{x^2}$ is 10.

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