Question

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Answer 1

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Answer 2

Answer:

$x^{2}+\frac{1}{x^{2}}= 14$ and $x^{2}-\frac{1}{x^{2}}= 10$

Step-by-step explanation:

In 4) Given $x= 2 + \sqrt{3}$

we need to find $x^{2}+\frac{1}{x^{2}}$ and $x^{2}-\frac{1}{x^{2}}$

If $x =2 + \sqrt{3}$

then to find $\frac{1}{x}$

Rationalize the $\frac{1}{x}= \frac{1}{2 + \sqrt{3}}$

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

$x= \frac{2-\sqrt{3}}{(2)^{2}-(\sqrt{3})^{2}}$

$x= \frac{2-\sqrt{3}}{4-3}$

$x= 2-\sqrt{3}$

Now, find $x^{2}+\frac{1}{x^{2}}$

Since, $(a+b)^{2} -2ab= a^{2} + b^{2}$

so,  $x^{2}+\frac{1}{x^{2}} = (x+\frac{1}{x})^{2}-2$

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

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

$= (4)^{2}-2$

$= 16-2$

$= 14$

Hence, $x^{2}+\frac{1}{x^{2}}= 14$

Now, find $x^{2}-\frac{1}{x^{2}}$

Since, $(a-b)^{2} +2ab= a^{2} + b^{2}$

so,  $x^{2}-\frac{1}{x^{2}} = (x+\frac{1}{x})^{2}+2$

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

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

$= 12-2$

$= 10$

Hence, $x^{2}-\frac{1}{x^{2}}= 10$