Question
If $\text{a}=\sqrt2+1,$ then write the value of $\text{a}-\frac{1}{\text{a}}.$
✓
Answer
Given that $\text{a}=\sqrt2+1,$
Hence $\frac{1}{\text{a}}$ is
given as $\frac{1}{\text{a}}=\frac{1}{\sqrt2+1}.$
we are asked to find $\text{a}-\frac{1}{\text{a}}$
We know that rationalization factor for $\sqrt2+1$ is $\sqrt2-1.$
We will multiply each side of the given expression $\frac{1}{\sqrt2+1}$ by $\sqrt2-1,$ to get $\frac{1}{\text{a}}=\frac{1}{\sqrt2-1}\times\frac{\sqrt2-1}{\sqrt2-1}$
$=\frac{\sqrt2-1}{\big(\sqrt2\big)^2-(1)^2}$
$=\frac{\sqrt2-1}{\sqrt2-1}$
$=\sqrt2-1$ Therefore, $\text{a}-\frac{1}{\text{a}}=\sqrt2+1-\big(\sqrt2-1\big)$
$=\sqrt2+1-\sqrt2+1$
$=2$ Hence the value of the given expresion is $2.$
Hence $\frac{1}{\text{a}}$ is
given as $\frac{1}{\text{a}}=\frac{1}{\sqrt2+1}.$
we are asked to find $\text{a}-\frac{1}{\text{a}}$
We know that rationalization factor for $\sqrt2+1$ is $\sqrt2-1.$
We will multiply each side of the given expression $\frac{1}{\sqrt2+1}$ by $\sqrt2-1,$ to get $\frac{1}{\text{a}}=\frac{1}{\sqrt2-1}\times\frac{\sqrt2-1}{\sqrt2-1}$
$=\frac{\sqrt2-1}{\big(\sqrt2\big)^2-(1)^2}$
$=\frac{\sqrt2-1}{\sqrt2-1}$
$=\sqrt2-1$ Therefore, $\text{a}-\frac{1}{\text{a}}=\sqrt2+1-\big(\sqrt2-1\big)$
$=\sqrt2+1-\sqrt2+1$
$=2$ Hence the value of the given expresion is $2.$
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