Question

if a=5+2√6 then find the value of a^2+1/a^2

Answer 1

Given,
a= 5+2√6

1/a= 1/(5+2√6)
multiplying with comgugates,
1/a = (5-2√6)/{(5+2√6)(5-2√6)

1/a = (5-2√6)/{5² -(2√6)²}

1/a = (5-2√6)/{25 - 24}

1/a = 5-2√6

Refer to attachment.
identities used
a²-b² = (a+b)(a-b)

....

Answer 2

Answer:

98 is the required value of $a^2+ (\frac{1}{a}) ^2$ .

Step-by-step explanation:

Explanation:

Given, a =$5 + 2\sqrt{6}$  and $a^2+ (\frac{1}{a}) ^2$ .

To find the value of $a^2+ (\frac{1}{a}) ^2$ , first, we rationalize $a^{2}$ .

Step 1:

Therefore, we have , $a^{2} = (5 +2\sqrt{6} )^{2}$   and $(\frac{1}{a})^{2} = (\frac{1}{ (5 +2\sqrt{6} )})^{2}$

First, we rationalize  the value of $a^{2} = (5+2\sqrt{6})^2$

⇒$a^{2} = ( 5+2\sqrt{6} .\frac{ 5-2\sqrt{6}}{ 5-2\sqrt{6} })^{2}$  

[Multiplying numerator and denomination with $5-2\sqrt{6}$ ]

⇒$a^{2}$ = $(\frac{(5 + 2\sqrt{6} )(5-2\sqrt{6}) }{5- 2\sqrt6x} })^2$ = $(\frac{5^2 - (2\sqrt{6} )^2}{5 -2\sqrt{6} } )^{2}$

⇒$a^2$ = $(\frac{25 - 24 )}{5 -2\sqrt{6} } )^{2}$ = $(\frac{1}{5 - 2\sqrt{6} } )^2$

Step 2:

From the question we have,  $a^2+ (\frac{1}{a}) ^2$  

Now, put the value of $a^2 and \ (\frac{1}{a} )^2$  we get,

⇒   $a^2+ (\frac{1}{a}) ^2$ =    $(\frac{1}{5 - 2\sqrt{6} } )^2 + (\frac{1}{5 + 2\sqrt{6} } )^2$ = $\frac{({5 + 2\sqrt{6} } )^2+({5 - 2\sqrt{6} } )^2 }{({5 - 2\sqrt{6} } )^2.({5 + 2\sqrt{6} } )^2}$

⇒  $a^2+ (\frac{1}{a}) ^2$ =  $\frac{(25 +24 +20\sqrt{6} + (25 + 24 - 20\sqrt{6} ) )}{(5^2-(2\sqrt{6} )^2)^2}$         [∴$a^2 - b^2 = (a-b)(a+b)$]

= $a^2+ (\frac{1}{a}) ^2$ = $\frac{98}{(25-24)^2}$     = 1

Final answer:

Hence, the  value of  $a^2+ (\frac{1}{a}) ^2$ is 98

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