If $x=8+3\sqrt{7},$ then what is value of $\sqrt{x} -\frac{1}{\sqrt{x}}$?
This question is somewhat different than I thought. I only know how to find the value when root is not given . Please help me.
2026-04-13 08:17:31.1776068251
If $x=8+3\sqrt{7},$ then what is value of $\sqrt{x} -\frac{1}{\sqrt{x}}$?
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Note that $\left ( \sqrt{x}- \frac{1}{\sqrt{x}} \right )^2 = x + \frac{1}{x} - 2$. Further note that $8+3 \sqrt{7} > 1$, so $\sqrt{x}-\frac{1}{\sqrt{x}} > 0$.
$$x=8+3 \sqrt{7} \implies \frac{1}{x} = \frac{8-3\sqrt{7}}{8^2 - (3\sqrt{7})^2} = \frac{8-3\sqrt{7}}{64-63} = 8-3\sqrt{7}.$$
Thus, $x+\frac{1}{x} = 16$, which implies $x + \frac{1}{x} - 2 = 14$. Thus, $\sqrt{x}- \frac{1}{\sqrt{x}} = \sqrt{14}$.