Question

If a=3+2√2 then find the value of a^1/2+a^-1/2.
a)√2
b)-√2
c)2√2
d)-2√2

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Answer 1

Given :  a = 3 + 2√2

To find :  $\sf{{a}^{{\frac{1}{2}}} + {a}^{{\frac{- 1}{2}}}}$

We can write this as -

$\longrightarrow\sf{ {\sqrt{a}} + {\dfrac{1}{ {\sqrt{a}} }} }$

Finding 1/a.

$\longrightarrow\sf{ {\dfrac{1}{ 3 + 2 {\sqrt{2}} }} }$

$\longrightarrow\sf{ {\dfrac{1}{ 3 + 2 {\sqrt{2}} }} \times  {\dfrac{3 - 2 {\sqrt{2}}}{ 3 - 2 {\sqrt{2}} }}}$

$\longrightarrow\sf{ {\dfrac{ 3 - 2 {\sqrt{2}} }{ (3)^2 - (2 {\sqrt{2}} )^2 }}}$

$\longrightarrow\sf{ {\dfrac{ 3 - 2 {\sqrt{2}} }{ 9 - 8 }}}$

$\longrightarrow\sf{ {\dfrac{1}{a}} = 3 - 2 {\sqrt{2}} }$

Now,

Squaring of √a + 1/√a.

$\sf{ \left( {\sqrt{a}} + {\dfrac{1}{ {\sqrt{a}} }} \right) ^2 = ( {\sqrt{a}} )^2 + \left( {\dfrac{1}{ {\sqrt{a}} }} \right) ^2 + 2 . {\sqrt{2}} . {\dfrac{1}{ {\sqrt{2}} }} }$

$\longrightarrow\sf{ a + {\dfrac{1}{a}} + 2}$

$\longrightarrow\sf{ 3 + 2 {\sqrt{2}} + 3 - 2 {\sqrt{2}} + 2}$

$\longrightarrow\sf{ 3 + 3 + 2}$

$\sf{ \left( {\sqrt{a}} + {\dfrac{1}{ {\sqrt{a}} }} \right) ^2 = 8}$

${\boxed{\sf{{\sqrt{a}} + {\dfrac{1}{ {\sqrt{a}} }} = {\sqrt{8}}}}}$