Question

Here's one more question from ch-6,class-8, maths,USA.

$if \: x = 2 + \sqrt{3} \: find \: {x }^{2} + \frac{1}{ {x}^{2} }$
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Answer 1

Step-by-step explanation:

x= 2 + √3

1/x= 1 /2+√3 =1 / 2+√3 * 2- √3 /2-√3 =2-√3

x^2 = 4+3+4√3= 7+4√3

1/ x^2 = 7- 4√3

x^2+1/x^2=14

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

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

Now, Consider

$\rm \: \dfrac{1}{x}$

$\rm \: = \: \dfrac{1}{2 + \sqrt{3} }$

On rationalizing the denominator, we get

$\rm \: = \: \dfrac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 + \sqrt{3} }$

We know,

$\boxed{ \rm{ \:(x + y)(x - y) = {x}^{2} - {y}^{2} \: }}$

So, using this identity, we get

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

$\rm \: = \: \dfrac{2 - \sqrt{3} }{4 - 3}$

$\rm \: = \: \dfrac{2 - \sqrt{3} }{1}$

$\rm \: = \: 2 - \sqrt{3}$

$\rm\implies \: \frac{1}{x} = \: 2 - \sqrt{3}$

Now, Consider

$\rm \: {x}^{2} + \dfrac{1}{ {x}^{2} }$

$\rm \: = \: {x}^{2} + {\bigg(\dfrac{1}{x} \bigg) }^{2}$

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

We know,

$\boxed{ \rm{ \: {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2}) \: }}$

So, using this identity, we get

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

$\rm \: = \: 2(4 + 3)$

$\rm \: = \: 2 \times 7$

$\rm \: = \: 14$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 14 \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$