If $${\left[ {a + \frac{1}{a}} \right]^2} - 2\left[ {a - \frac{1}{a}} \right] = 12,$$ then which of the following is a value of 'a'?
A. -8 + √3
B. -8 - √3
C. -8 + √5
D. None of these
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & {\left[ {a + \frac{1}{a}} \right]^2} - 2\left[ {a - \frac{1}{a}} \right] = 12 \cr & {\left[ {a - \frac{1}{a}} \right]^2} + 4 - 2\left[ {a - \frac{1}{a}} \right] = 12 \cr & {\left[ {a - \frac{1}{a}} \right]^2} - 2\left[ {a - \frac{1}{a}} \right] - 8 = 0 \cr & {\text{Let }}a - \frac{1}{a} = x \cr & \therefore \,{x^2} - 2x - 8 = 0 \cr & {x^2} - 4x + 2x - 8 = 0 \cr & \left( {x - 4} \right)\left( {x + 2} \right) = 0 \cr & x = 4,\,x = - 2 \cr & a - \frac{1}{a} = - 2,\,a - \frac{1}{a} = 4 \cr & \therefore \,a + \frac{1}{a} = \sqrt {{4^2} + 4} \cr & a + \frac{1}{a} = 2\sqrt 5 \cr & a - \frac{1}{a} = 4 \cr & \therefore \,2a = 2\sqrt 5 + 4 \cr & a = 2 + \sqrt 5 \cr} $$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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