Question

ANSWER CORRECTLY WITHOUT SPAMING ​

Answer 1

$\large \bf \clubs \:  Given :-$

$\sf \mathtt x = 5 - 2 \sqrt{6}$

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$\large \bf \clubs \:  To \: Find :-$

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

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$\large \bf \clubs \:  Solution :-$

We Have,

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

Adding and Subtracting 2 :

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

Putting Value of x :

$= \sf {\bigg(5 - 2 \sqrt{6} + \dfrac{1}{5 - 2 \sqrt{6} } \bigg)}^{2} - 2\\ \\ \sf = {\bigg(5 - 2 \sqrt{6} + \frac{1}{5 - 2 \sqrt{6}} \times \frac{5 + 2 \sqrt{6} }{5 + 2 \sqrt{6} } \bigg)}^{2} - 2 \\ \\ \sf = { \bigg(5 - 2 \sqrt{6} + \frac{5 + 2 \sqrt{6} }{25 - 24} \bigg)}^{2} - 2 \\ \\ \sf = (5 - \cancel{2\sqrt{6}} + 5 + \cancel{ 2 \sqrt{6}} ) {}^{2} - 2 \\ \\ \sf = {(10)}^{2} - 2 \\ \\ = 100 - 2 \\ \\ = 98$

Hence,

$\qquad\Large\underline{\underline{ \pink{\bf {x}^{2} + \dfrac{1}{ {x}^{2}}=98 }}}$

$\Large\red{\mathfrak{  \text{W}hich \:\:is\:\: the\:\: required} }\\ \LARGE \red{\mathfrak{ \text{ A}nswer.}}$

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Answer 2

$\overline{ \underline{ \boxed{ \sf QUESTION \: \darr}}}$

• If x = 5 - 2√6, find x² + 1/x²

$\overline{ \underline{ \boxed{ \sf IDENTITY \: \darr}}}$

$\star \:\orange{\underline\green{{ \boxed{ \red{ \mathfrak{ {a}^{2} + {b}^{2} = {(a + b)}^{2} - 2ab}}}}}}$

$\overline{ \underline{ \boxed{ \sf CONCEPT \: \darr}}}$

Rarionalsation :

• The process of multiplying a surd by another surd to get a rational number is called rationalisation.

• Each surd is called rationalising factor of another surd.

Rule to Rationalise the denominator of an expression :

• Multiply and divide the numerator and denominator of the given expression by rationalising factor of its denominator and simplify.

$\overline{ \underline{ \boxed{ \sf SOLUTION \: \darr}}}$

Given that :

x = 5 - 2√6

So,

1/x must be 1/(5 - 2√6) by putting the value of x in the denominator

Now to rationalise the denominator :

$\sf \looparrowright \frac{1}{x} = \frac{1}{5 - 2 \sqrt{6} }$

$\sf \looparrowright \frac{1}{x} = \frac{1(5 + 2 \sqrt{6} )}{(5 - 2 \sqrt{6})(5 + 2 \sqrt{6}) }$

Now applying identity for the denominator :

(a - b)(a + b) = a² - b²

Here,

a → 5

b → 2√6

$\sf \looparrowright\frac{1}{x} = \frac{5 + 2 \sqrt{6} }{ {(5)}^{2} - {(2 \sqrt{6} )}^{2} }$

$\sf \looparrowright\frac{1}{x} = \frac{5 + 2 \sqrt{6} }{25 - 24}$

$\star \: \red{\underline{ \small{ \purple{\boxed{ \frak{ \green{\frac{1}{x} = 5 + 2 \sqrt{6}}}}}}}}$

Now adding x with 1/x we get

$\sf \rarr x + \frac{1}{x}$

Putting their values we get

$\sf \rarr5 + 2 \sqrt{6} + 5 - 2 \sqrt{6}$

$\sf \rarr5 + \cancel{2 \sqrt{6} } + 5 - \cancel{2 \sqrt{6} }$

$\star \: \underline{ \boxed{\mathfrak{x + \frac{1}{x} = 10}}}$

Now we need to find x² + 1/x²

Applying the identity we get

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

$\sf \leadsto {x}^{2} + \frac{1}{ {x}^{2} } = {(10)}^{2} - 2$

$\sf \leadsto {x}^{2} + \frac{1}{ {x}^{2} } = 100 - 2$

$\leadsto \underline{ \boxed{ \mathfrak{{x}^{2} + \frac{1}{ {x}^{2} } = 98}}}$

• Hence, the answer is 98

$\overline{ \underline{ \boxed{ \sf LEARN \: MORE \: \darr}}}$

$\star \: \underline{ \boxed{ \mathfrak{ {(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2} }}}$

$\star \: \underline{ \boxed{ \mathfrak{ {(a - b)}^{2} = {a}^{2} - 2ab + {b}^{2} }}}$

$\star \: \underline{ \boxed{ \mathfrak{ {a}^{2} + {b}^{2} = {(a + b)}^{2} - 2ab }}}$

$\star \: \underline{ \boxed{ \mathfrak{ {a}^{2} + {b}^{2} = {(a - b)}^{2} + 2ab}}}$

$\star \: \underline{ \boxed{ \mathfrak{ {(a + b)}^{2} + {(a - b)}^{2} = 2( {a}^{2} + {b}^{2} )}}}$

$\star \: \underline{ \boxed{ \mathfrak{ {(a + b)}^{2} - {(a - b)}^{2} = 4ab }}}$

$\star \: \underline{ \boxed{ \mathfrak{ {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca) }}}$

$\star \: \underline{ \boxed{ \mathfrak{ {(a + b - c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc - ca) }}}$