If $$x + \frac{1}{x} = - 14,$$ and x < -1 what will be the value of $${x^2} - \frac{1}{{{x^2}}} = ?$$
A. -112√3
B. 112√3
C. -140√2
D. 140√2
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & x + \frac{1}{x} = - 14 \cr & x - \frac{1}{x} = \sqrt {{{\left( { - 14} \right)}^2} - 4} \cr & = \sqrt {196 - 4} \cr & = \sqrt {192} \cr & = 8\sqrt 3 \cr & \therefore \,x < - 1 \cr & \left( {x + \frac{1}{x}} \right)\left( {x - \frac{1}{x}} \right) = - 14x\left( {8\sqrt 3 } \right) \cr & {x^2} - \frac{1}{{{x^2}}} = + 112\sqrt 3 \cr} $$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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