Question
If $\text{x}=3+\sqrt8,$ find the value of $\text{x}^2+\frac{1}{\text{x}^2}.$
✓
Answer
We know that $\text{x}^2+\frac{1}{\text{x}^2}=\Big(\text{x}+\frac{1}{\text{x}^2}\Big)^2-2.$ we have to find the value of $\text{x}^2+\frac{1}{\text{x}^2}.$ As $\text{x}=3+\sqrt8$ therefore,$\frac{1}{\text{x}}=\frac{1}{3+\sqrt8}$
We know that rationalization factor for $3+\sqrt8$ is $3-\sqrt8.$ We will multiply numerator and denominator of the given expression $\frac{1}{3+\sqrt8}$ by $3-\sqrt8,$ to get.$\frac{1}{\text{x}}=\frac{1}{3+\sqrt8}\times\frac{3-\sqrt8}{3-\sqrt8}$
$=\frac{3-\sqrt8}{(3)^2-\big(\sqrt8\big)^2}$
$=\frac{3-\sqrt8}{9-8}$
$=3-\sqrt8$
Putting the value of x and $\frac{1}{\text{x}},$ we get$\text{x}^2+\frac{1}{\text{x}^2}=\big(3+\sqrt8+3-\sqrt8\big)^2-2$
$=(6)^2-2$
$=36-2$
$=34$
Hence the given expression is simplified to 34.
We know that rationalization factor for $3+\sqrt8$ is $3-\sqrt8.$ We will multiply numerator and denominator of the given expression $\frac{1}{3+\sqrt8}$ by $3-\sqrt8,$ to get.$\frac{1}{\text{x}}=\frac{1}{3+\sqrt8}\times\frac{3-\sqrt8}{3-\sqrt8}$
$=\frac{3-\sqrt8}{(3)^2-\big(\sqrt8\big)^2}$
$=\frac{3-\sqrt8}{9-8}$
$=3-\sqrt8$
Putting the value of x and $\frac{1}{\text{x}},$ we get$\text{x}^2+\frac{1}{\text{x}^2}=\big(3+\sqrt8+3-\sqrt8\big)^2-2$
$=(6)^2-2$
$=36-2$
$=34$
Hence the given expression is simplified to 34.
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