Question

If = 2 + √3, find the value of  x^2 + 1/ x^ 2 . [Hint: use the identity ( a+b ) ^2 ]


Pls answer immediately.

Whoever answers fast and correct will be marked as the brainliest

Answer 1

CorrecT QuestioN :

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

$\sf{Find\:the\: value\:of\:x^2+\frac{1}{x^2}}$

[ Hints : Use the identity (a+b)² ]

SolutioN :

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

✪ Now find the value of $\frac{1}{x}$✪

$\frac{1}{x} = \frac{1}{2 + \sqrt{3} } \\ \: \: \: \: \: = \frac{2 - \sqrt{3} }{(2 + \sqrt{3} )(2 - \sqrt{3} )}$

★Using the identity a²-b²=(a+b)(a-b)★

$\: \: \: \: \: = \frac{2 - \sqrt{3} }{ {2}^{2} - { (\sqrt{3}) }^{2} } \\ \: \: \: \: \: = \frac{2 - \sqrt{3} }{4 - 3} \\ \: \: \: \: \: = \frac{2 - \sqrt{3} }{1} \\ \: \: \: \: \: = 2 - \sqrt{3}$

✪ $\sf{Now\:find\:the\: value\:of\:x^2+\frac{1}{x^2}}$✪

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

★ Putting values :-

•x = 2+√3

•1/x = 2-√3 ★

$= {(2 + \sqrt{3} + 2 - \sqrt{3}) }^{2} - 2 \\ = {4}^{2} - 2 \\ = 16 - 2 \\ = 14(Answer)$

Answer 2

GIVEN:

★$x = 2+\sqrt{3}$

TO FIND:

.

★$x^{2}+\dfrac{1}{x^{2}}$

CONCEPT USED:

★We would first find $\dfrac{1}{x}$

★Then we would find $x^{2}+\dfrac{1}{x^{2}}$

ANSWER:

Given,

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

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

On rationalising the denominator,

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

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

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

=>$\dfrac{1}{x}=\dfrac{2-\sqrt{3}}{4-3}$

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

.°. $\dfrac{1}{x}=2-\sqrt{3}$

______________________________________

Now,

=$x^{2}+\dfrac{1}{x^{2}}$

=$x^{2}+\dfrac{1}{x^{2}}-2×\dfrac{1}{x}×x+2×\dfrac{1}{x}{x}$

$\large\purple{\boxed{(a+b) ^{2}=a^{2}+b^{2}+2ab}}$

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

On substituting the values,

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

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

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

=>$x^{2}+\dfrac{1}{x^{2}}= 12+2$

.°.$x^{2}+\dfrac{1}{x^{2}}=14$

$\huge\orange{\boxed{.°.x^{2}+\dfrac{1}{x^{2}}=14}}$