If$$\ x^{2}+\frac{1}{x^{2}}=\frac{7}{4}$$ for x > 0, then what is the value of x +Â $$\frac{1}{x}$$?
Let,
$$(x + \frac{1}{x})^{2}$$ =Â $$\ x^{2}+\frac{1}{x^{2}} + 2(x)(\frac{1}{x})$$
$$(x + \frac{1}{x})^{2}$$ =Â $$\frac{7}{4} + 2$$
$$(x + \frac{1}{x})^{2}$$ =Â $$\frac{15}{4}$$
$$(x + \frac{1}{x})$$ =Â $$\frac{\sqrt{15}}{2}$$
Hence, option B is the correct answer.
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