If $$a = 2 + \sqrt 3 {\text{,}}$$ then the value of $$\left( {{a^2} + \frac{1}{{{a^2}}}} \right) = \,?$$
A. 12
B. 14
C. 16
D. 10
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & a = 2 + \sqrt 3 \cr & \Rightarrow {a^2} = {\left( {2 + \sqrt 3 } \right)^2} \cr & \Rightarrow {a^2} = 4 + 3 + 4\sqrt 3 \cr & \Rightarrow {a^2} = 7 + 4\sqrt 3 \cr & \frac{1}{{{a^2}}} = \frac{1}{{7 + 4\sqrt 3 }} \cr & \Rightarrow \frac{1}{{{a^2}}} = \frac{{7 - 4\sqrt 3 }}{{\left( {7 + 4\sqrt 3 } \right)\left( {7 - 4\sqrt 3 } \right)}} \cr & \Rightarrow \frac{1}{{{a^2}}} = \frac{{7 - 4\sqrt 3 }}{1} \cr & \Rightarrow \frac{1}{{{a^2}}} = 7 - 4\sqrt 3 \cr & \therefore {a^2} + \frac{1}{{{a^2}}} \cr & = 7 + 4\sqrt 3 + 7 - 4\sqrt 3 \cr & = 14 \cr} $$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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