If $$x = 3 + \sqrt 8 ,$$ then $${x^2} + \frac{1}{{{x^2}}}$$ is equal to = ?
A. 30
B. 34
C. 36
D. 38
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & \because x = 3 + \sqrt 8 \cr & \Rightarrow {x^2} = {\left( {3 + \sqrt 8 } \right)^2} \cr & \Rightarrow {x^2} = {3^2} + {\left( {\sqrt 8 } \right)^2} + 2 \times 3 \times \sqrt 8 \cr & \Rightarrow {x^2} = 9 + 8 + 6\sqrt 8 \cr & \Rightarrow {x^2} = 17 + 12\sqrt 2 \cr} $$$$\therefore {x^2} + \frac{1}{{{x^2}}}$$
$$ = \left( {17 + 12\sqrt 2 } \right)$$ $$ + \frac{1}{{\left( {17 + 12\sqrt 2 } \right)}}$$ $$ \times \frac{{\left( {17 - 12\sqrt 2 } \right)}}{{\left( {17 - 12\sqrt 2 } \right)}}$$
$$\eqalign{ & = \left( {17 + 12\sqrt 2 } \right) + \frac{{\left( {17 - 12\sqrt 2 } \right)}}{{289 - 288}} \cr & = \left( {17 + 12\sqrt 2 } \right) + \left( {17 - 12\sqrt 2 } \right) \cr & = 17 + 17 \cr & = 34 \cr} $$
Related Questions on Square Root and Cube Root
The least perfect square, which is divisible by each of 21, 36 and 66 is:
A. 213444
B. 214344
C. 214434
D. 231444
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