Question

If x = 3 $-$ $\sqrt{18}$ then find the value of $\sqrt{x}-\frac{1}{\sqrt{x}}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}x=3-\sqrt{8}=3-2\sqrt{2} \\ \text{So},\frac{1}{x}=\frac{1}{3-2\sqrt{2}}=\frac{1}{3-2\sqrt{2}}\times \frac{3+2\sqrt{2}}{3+2\sqrt{2}} \\ =\frac{3+2\sqrt{2}}{{\left(3\right)}^2-{\left(2\sqrt{2}\right)}^2} \\ =\frac{3+2\sqrt{2}}{9-8} \\ =3+2\sqrt{2} \\ \text{So},x+\frac{1}{x}=3-2\sqrt{2}+3+2\sqrt{2}=6 \\ \mathrm{We}\mathrm{know}\mathrm{that} \\ {\left(\sqrt{\mathrm{x}}-\frac{1}{\sqrt{\mathrm{x}}}\right)}^2=x+\frac{1}{x}-2 \\ \Rightarrow {\left(\sqrt{\mathrm{x}}-\frac{1}{\sqrt{\mathrm{x}}}\right)}^2=6-2 \\ \Rightarrow {\left(\sqrt{\mathrm{x}}-\frac{1}{\sqrt{\mathrm{x}}}\right)}^2=4 \\ \mathrm{Taking}\mathrm{square}\mathrm{root}\mathrm{on}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}: \\ \left(\sqrt{\mathrm{x}}-\frac{1}{\sqrt{\mathrm{x}}}\right)=\pm 2 \end{gathered}$$