Question
If $\text{x}=3+2\sqrt2,$ than find the value of $\sqrt{\text{x}}-\frac{1}{\sqrt{\text{x}}}.$
✓
Answer
We are asked to simplify $\text{x}=3+2\sqrt2.$ It can be written in the form $(\text{a}+\text{b})^2=\text{a}^2+\text{b}^2+2\text{ab}$ as $\sqrt{\text{x}}=\sqrt{3+2\sqrt2}$
$=\sqrt{2+1+2\times1\times\sqrt2}$
$=\sqrt{\big(\sqrt2\big)^2+(1)^2+2\times1\times\sqrt2}$
$=\sqrt{\big(\sqrt2+1\big)^2}$
$=\sqrt2+1$ Therefore, $\frac{1}{\sqrt{\text{x}}}=\frac{1}{\sqrt2+1}$
We know that rationalization factor for $\sqrt2+1$ is $\sqrt2-1.$
We will multiply numerator and denominator of the given expression $\frac{1}{\sqrt2+1}$ by $\sqrt2-1,$ to get $\frac{1}{\sqrt2+1}\times\frac{\sqrt2-1}{\sqrt2-1}=\frac{\sqrt2-1}{\big(\sqrt2\big)^2-(1)^2}$
$=\frac{\sqrt2-1}{2-1}$
$=\sqrt2-1$ Hence, $\sqrt{\text{x}}-\frac{1}{\sqrt{\text{x}}}=\sqrt2+1-\big(\sqrt2-1\big)$
$\sqrt2+1-\sqrt2+1$
$=2$ Therefore, value of given expression is 2.
$=\sqrt{2+1+2\times1\times\sqrt2}$
$=\sqrt{\big(\sqrt2\big)^2+(1)^2+2\times1\times\sqrt2}$
$=\sqrt{\big(\sqrt2+1\big)^2}$
$=\sqrt2+1$ Therefore, $\frac{1}{\sqrt{\text{x}}}=\frac{1}{\sqrt2+1}$
We know that rationalization factor for $\sqrt2+1$ is $\sqrt2-1.$
We will multiply numerator and denominator of the given expression $\frac{1}{\sqrt2+1}$ by $\sqrt2-1,$ to get $\frac{1}{\sqrt2+1}\times\frac{\sqrt2-1}{\sqrt2-1}=\frac{\sqrt2-1}{\big(\sqrt2\big)^2-(1)^2}$
$=\frac{\sqrt2-1}{2-1}$
$=\sqrt2-1$ Hence, $\sqrt{\text{x}}-\frac{1}{\sqrt{\text{x}}}=\sqrt2+1-\big(\sqrt2-1\big)$
$\sqrt2+1-\sqrt2+1$
$=2$ Therefore, value of given expression is 2.
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