Question
If $\text{x}=2+\sqrt3,$ find the value of $\text{x}+\frac{1}{\text{x}}.$
✓
Answer
Given that $\text{x}=2+\sqrt3,$ Hence $\frac{1}{\text{x}}$ is given as
$\frac{1}{\text{x}}=\frac{1}{2+\sqrt3}.$ we are asked to find $\text{x}+\frac{1}{\text{x}}$
We know that rationalization factor for $2+\sqrt3$ is $2-\sqrt3.$ We will multiply each side of the given expression $\frac{1}{2+\sqrt3}$ by $2-\sqrt3,$ to get
$\frac{1}{\text{x}}=\frac{1}{2+\sqrt3}\times\frac{2-\sqrt3}{2-\sqrt3}$
$=\frac{2-\sqrt3}{(2)^2-\big(\sqrt3\big)^2}$
$=\frac{2-\sqrt3}{4-3}$
$=2-\sqrt3$
Therefore,
$\text{x}+\frac{1}{\text{x}}=2+\sqrt3+2-\sqrt3$
$=4$
Hence the value of the given expresion is 4.
$\frac{1}{\text{x}}=\frac{1}{2+\sqrt3}.$ we are asked to find $\text{x}+\frac{1}{\text{x}}$
We know that rationalization factor for $2+\sqrt3$ is $2-\sqrt3.$ We will multiply each side of the given expression $\frac{1}{2+\sqrt3}$ by $2-\sqrt3,$ to get
$\frac{1}{\text{x}}=\frac{1}{2+\sqrt3}\times\frac{2-\sqrt3}{2-\sqrt3}$
$=\frac{2-\sqrt3}{(2)^2-\big(\sqrt3\big)^2}$
$=\frac{2-\sqrt3}{4-3}$
$=2-\sqrt3$
Therefore,
$\text{x}+\frac{1}{\text{x}}=2+\sqrt3+2-\sqrt3$
$=4$
Hence the value of the given expresion is 4.
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