Question : If $x^{2} -3x +1=0$, then the value of $\frac{\left(x^4+\frac{1}{x^2}\right)}{\left(x^2+5 x+1\right)}$ is:
Option 1: $\frac{9}{4}$
Option 2: $\frac{27}{8}$
Option 3: $\frac{5}{2}$
Option 4: $2$
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Correct Answer: $\frac{9}{4}$
Solution :
Given:
$x^{2} -3x +1=0$
$⇒x^{2} +1 = 3x$
Dividing $x$ on both sides, we get,
$⇒x + \frac{1}{x} = 3$
Cubing both sides, we get,
$⇒(x+\frac{1}{x})^3=3^3$
$⇒x^3+\frac{1}{x^3}+3×x×\frac{1}{x}(x+\frac{1}{x})=27$
$⇒x^3+\frac{1}{x^3}+3×3=27$
$⇒x^3+\frac{1}{x^3}=18$
Now,
$ \frac{x^4 +\frac{1}{x^2}}{x^2 + 5x + 1}$
$= \frac{x(x^3 +\frac{1}{x^3})}{x(x + 5 + \frac{1}{x})}$
$= \frac{x^3 +\frac{1}{x^3}}{x + 5 + \frac{1}{x}}$
$= \frac{18}{3+5}$
$= \frac{18}{8}$
$= \frac{9}{4}$
Hence, the correct answer is $\frac{9}{4}$.
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