If $$x = 3 + 2\sqrt 2 {\text{,}}$$ then the value of $${x^2}{\text{ + }}\frac{1}{{{x^2}}}{\text{ is?}}$$
A. 36
B. 30
C. 32
D. 34
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & {\text{ }}x = 3 + 2\sqrt 2 \cr & \Rightarrow {x^2} = {\left( {3 + 2\sqrt 2 } \right)^2} \cr & \left( {{\text{Squaring both sides}}} \right) \cr & \Rightarrow {x^2} = 9 + 8 + 12\sqrt 2 \cr & \Rightarrow {x^2} = 17 + 12\sqrt 2 \cr & \Rightarrow \frac{1}{{{x^2}}} = \frac{1}{{17 + 12\sqrt 2 }} \times \frac{{17 - 12\sqrt 2 }}{{17 - 12\sqrt 2 }} \cr & \Rightarrow \frac{1}{{{x^2}}} = 17 - 12\sqrt 2 \cr & \therefore {\text{ }}{x^2}{\text{ + }}\frac{1}{{{x^2}}} \cr & = 17 + 12\sqrt 2 + 17 - 12\sqrt 2 \cr & = 34 \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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