Question 57

If $$x^4+\frac{1}{x^4}=14159$$, then the value of $$x+\frac{1}{x}$$ is:

Solution

Given,  $$x^4+\frac{1}{x^4}=14159$$

$$\Rightarrow$$  $$x^4+\frac{1}{x^4}+2=14159+2$$

$$\Rightarrow$$  $$\left(x^2+\frac{1}{x^2}\right)^2=14161$$

$$\Rightarrow$$  $$\left(x^2+\frac{1}{x^2}\right)^2=\left(119\right)^2$$

$$\Rightarrow$$  $$x^2+\frac{1}{x^2}=119$$

$$\Rightarrow$$  $$x^2+\frac{1}{x^2}+2=119+2$$

$$\Rightarrow$$  $$\left(x+\frac{1}{x}\right)^2=121$$

$$\Rightarrow$$   $$x+\frac{1}{x}=11$$

Hence, the correct answer is Option A


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