Question

x=2-√5 find x^2 + 1/x^2​

Answer 1

Question:

If $x=2- \sqrt{5}$ find $x^{2} +\dfrac{1}{x^{2} }$.

To find:

$\bigstar$ To find the value of $x^{2} +\dfrac{1}{x^{2} }$.

Answer:

The value of $\bold{\underline{\:x^{2} +\dfrac{1}{x^{2} }\:is\:18\:}}$

Given:

$\bigstar \: x=2-\sqrt{5}$

Step-by-step explanation:

$\hookrightarrow \: x=2-\sqrt{5}$

$\hookrightarrow \dfrac{1}{x} =\dfrac{1}{2-\sqrt{5} }$

Here Rationalizing the denominator,

$\hookrightarrow \dfrac{1}{x} =\dfrac{2+\sqrt{5} }{(2-\sqrt{5})(2+\sqrt{5}) }$

It's of the form, $\underline{a^{2} -b^{2} =(a+b)(a-b)}$

$\hookrightarrow \dfrac{1}{x} =\dfrac{2+\sqrt{5} }{(2)^{2} -(\sqrt{5}) ^{2} }$

$\hookrightarrow \dfrac{1}{x} =\dfrac{2+\sqrt{5} }{4-5}$

$\hookrightarrow\boxed{ \dfrac{1}{x} =2+\sqrt{5} }$

Now, find the value of $x^{2} +\dfrac{1}{x^{2} }$

Here Substituting the value of $x$ and $\dfrac{1}{x}$ ,

$\implies (2-\sqrt{5} )^{2} +(2+\sqrt{5} )^{2}$

It's of the form $(a+b)^{2}$ and $(a-b)^{2}$

$\implies \big[ (2)^{2} -2 \times 2 \times \sqrt{5} +(\sqrt{5} )^{2}\big] +\big[ (2)^{2} +2 \times 2 \times \sqrt{5} +(\sqrt{5} )^{2}\big]$

$\implies \big[4-4\sqrt{5} +5\big]+\big[4+4\sqrt{5} +5\big]$

$\implies 9 -\underline{\:4 \sqrt{5}\:} +9+\underline{\:4\sqrt{5} \:}$

$\implies \boxed{x^{2} +\dfrac{1}{x^{2} }=18}$

$\therefore$ The value of $\bold{\underline{\:x^{2} +\dfrac{1}{x^{2} }\:is\:18\:}}$.