Question

if x is equal to 3 minus 2 root 2 then find the value of x square minus one upon x square

Answer 1

$\bold{x^{2}-\frac{1}{x^{2}}=-24 \sqrt{2}}$ If $\bold{x=3-2 \sqrt{2}}$

Given:

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

To find:  

$x^{2}-\frac{1}{x^{2}}=?$

Solution:

The value of $x=3-2 \sqrt{2}$ is given in the question and by finding the value of $x^2$ and $\frac{1}{x^2}$and substituting the value of each term. Then the value of $x^{2}-\frac{1}{x^{2}}$ is obtained.

Let us consider $\frac{1}{x}=\frac{1}{3-2 \sqrt{2}}$

Rationalising the denominator gives  

$\begin{array}{l}{=\frac{1}{3-2 \sqrt{2}} \times \frac{3+2 \sqrt{2}}{3+2 \sqrt{2}}} \\ {=\frac{3+2 \sqrt{2}}{3^{2}-(2 \sqrt{2})^{2}}} \\ {=\frac{3+2 \sqrt{2}}{9-8}} \\ {=3+2 \sqrt{2}}\end{array}$

Now, $x^{2}-\frac{1}{x^{2}}=(3-2 \sqrt{2})^{2}-(3+2 \sqrt{2})^{2}$

Applying the algebraic formula $(a-b)^2 =a^2+b^2-2ab$

$=3^{2}+(2 \sqrt{2})^{2}-2 \times 3 \times 2 \sqrt{2}-\left(3^{2}+(2 \sqrt{2})^{2}+2 \times 3 \times 2 \sqrt{2}\right)$

Simplifying the formula and terms we get

$\begin{array}{l}{=9+8-12 \sqrt{2}-9-8-12 \sqrt{2} |} \\ {=-24 \sqrt{2}}\end{array}$

Therefore, $\bold{x^{2}-\frac{1}{x^{2}}=-24 \sqrt{2}}$  if $\bold{x=3-2 \sqrt{2}.}$

Answer 2

Step-by-step explanation:

Hope it helps mark brainliest