Question

If a=3+2√2 find the value of a²-1​/a²​

Answer 1

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Answer 2

Question:

If $a = 3 +2 \sqrt{2}$ find the value of ${a}^{2} - \dfrac{1}{ {a}^{2} }$

To find:

$\star \: {a}^{2} - \dfrac{1}{ {a}^{2} } = \: ?$

Answer:

$The \: value \: of \: \underline{ \: \underline{ \: \sf \pink{ \bf{{a}^{2} - \frac{1}{ {a}^{2} }\: is\: 0} } \: }\: }$

Given:

$\star \: a = 3 +2 \sqrt{2}$

Step-by-step explanation:

$\: a = 3 +2 \sqrt{2}$

$\dfrac{1}{ a } = \dfrac{1}{3 + 2 \sqrt{2} }$

Now rationalising the denominator,

$\rightsquigarrow \dfrac{1}{ a } = \dfrac{3 - 2 \sqrt{2} }{(3 + 2 \sqrt{2})(3 - 2 \sqrt{2}) }$

We know that,

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

$\rightsquigarrow \dfrac{1}{ a } = \dfrac{3 - 2 \sqrt{2} }{ {(3)}^{2} - {(2 \sqrt{2}) }^{2} }$

$\rightsquigarrow \dfrac{1}{ a } = \dfrac{3 - 2 \sqrt{2} }{ 9 - 8 }$

$\rightsquigarrow \boxed{\dfrac{1}{ a } = 3 - 2 \sqrt{2} }$

Method 1 :-

Now,

$\bigstar \: \bf{a}^{2} - \dfrac{1}{ {a}^{2} }$

$\implies {a}^{2} - \dfrac{1}{ {a}^{2} } = (a - \frac{1}{a}) (a + \frac{1}{a} )$

Now substituting the values, $a \: and \: \frac{1}{a}$

$= \big( \bold{3 + 2 \sqrt{2} - 3 - 2 \sqrt{2}} \: \big) \big(3 \bold{+ 2 \sqrt{2}} + 3 \bold{- 2 \sqrt{2}} \: \big)$

$= (0)(6)$

$= \boxed{\bf{0}}$

$\longmapsto \boxed{ {a}^{2} - \dfrac{1}{ {a}^{2} } = 0}$

$\therefore$The value of $\underline{ \: \bf{{a}^{2} - \frac{1}{ {a}^{2} }\: is\: 0}\:}$

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Method 2 :-

$\bigstar \: \bf{a}^{2} - \dfrac{1}{ {a}^{2} }$

$={(a)}^{2} - {\big( \dfrac{1}{a}\big)}^{2}$

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

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

$= ({3}^{2}+2 \times 3 \times 2 \sqrt{2}+8)-({3}^{2}-2 \times 3 \times 2 \sqrt{2}+8)$

$=9+8 \sqrt{2}+8-9-8 \sqrt{2}+8$

$= \boxed{\bold{0}}$

$\hookrightarrow \boxed{ {a}^{2} - \dfrac{1}{ {a}^{2} } = 0}$

$\therefore$The value of $\underline{ \: \bf{{a}^{2} - \frac{1}{ {a}^{2} }\: is\: 0}\:}$

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Formula used:

$\bigstar {a}^{2} - {b}^{2} = (a - b)(a + b)$

$\bigstar {a+b}^{2} ={a}^{2}+2ab+{b}^{2}$

$\bigstar {a-b}^{2} ={a}^{2}-2ab+{b}^{2}$