If\ x^{2}+\frac{1}{x^{2}}=\frac{7}{4} for x > 0, then what is the value of x + \frac{1}{x}?

Question 134

If$$\ x^{2}+\frac{1}{x^{2}}=\frac{7}{4}$$ for x > 0, then what is the value of x + $$\frac{1}{x}$$?

Solution

Let,

$$(x + \frac{1}{x})^{2}$$ = $$\ x^{2}+\frac{1}{x^{2}} + 2(x)(\frac{1}{x})$$

$$(x + \frac{1}{x})^{2}$$ = $$\frac{7}{4} + 2$$

$$(x + \frac{1}{x})^{2}$$ = $$\frac{15}{4}$$

$$(x + \frac{1}{x})$$ = $$\frac{\sqrt{15}}{2}$$

Hence, option B is the correct answer.

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