If x2 - 3x + 1 = 0, then the value of $$\frac{{\left( {{x^4} + \frac{1}{{{x^2}}}} \right)}}{{\left( {{x^2} + 5x + 1} \right)}}$$ is:
A. $$2$$
B. $$\frac{5}{2}$$
C. $$\frac{9}{4}$$
D. $$\frac{{27}}{8}$$
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & {x^2} - 3x + 1 = 0 \cr & {x^2} + 1 = 3x \cr & x + \frac{1}{x} = 3 \cr & \Rightarrow \frac{{\left( {{x^4} + \frac{1}{{{x^2}}}} \right)}}{{\left( {{x^2} + 5x + 1} \right)}} \cr & = \frac{{x\left( {{x^3} + \frac{1}{{{x^3}}}} \right)}}{{\left( {{x^2} + 1 + 5x} \right)}} \cr & = \frac{{x\left[ {{3^3} - 3 \times 3} \right]}}{{\left( {3x + 5x} \right)}} \cr & = \frac{{x\left[ {18} \right]}}{{\left( {8x} \right)}} \cr & = \frac{9}{4} \cr} $$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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