If $$x = \sqrt a + \frac{1}{{\sqrt a }}{\text{,}}$$ $$y = \sqrt a - \frac{1}{{\sqrt a }}{\text{,}}$$ $$\left( {a > 0} \right)$$ then the value of x4 + y4 - 2x2y2 is?
A. 16
B. 20
C. 10
D. 5
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & {\text{ }}x = \sqrt a + \frac{1}{{\sqrt a }} \cr & {\text{ }}y = \sqrt a - \frac{1}{{\sqrt a }} \cr & {\text{Put }}a = 4 \cr & x = 2 + \frac{1}{2} = \frac{5}{2} \cr & y = 2 - \frac{1}{2} = \frac{3}{2} \cr & {\text{Then, }}{x^4} + {y^4} - 2{x^2}{y^2}{\text{ }} \cr & = {\left( {{x^2} - {y^2}} \right)^2} \cr & = {\left( {\frac{{25}}{4} - \frac{9}{4}} \right)^2} \cr & = {\text{ 16}} \cr} $$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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