Question
If $x =(7+4 \sqrt{3})$, find the value of $x^2+\frac{1}{x^2}$
✓
Answer
$ x^2+\frac{1}{x^2}$
$\left(x^2+\frac{1}{x^2}\right)=\left(x+\frac{1}{x}\right)^2-2 \dots..... (1) $
We first find out $x+\frac{1}{x}$
$x+\frac{1}{x}=(7+4 \sqrt{3})^x+\frac{1}{(7+4 \sqrt{3})}$
$=\frac{(7+4 \sqrt{3})^2+1}{(7+4 \sqrt{3})}$
$=\frac{49+48+56 \sqrt{3}+1}{(7+4 \sqrt{3})}$
$=\frac{98+56 \sqrt{3}}{(7+4 \sqrt{3})}$
$=\frac{14(7+4 \sqrt{3})}{(7+4 \sqrt{3})}$
$=14 $
substitutingin $(1)$
$ \left(x^2+\frac{1}{x}\right)^2=\left(x+\frac{1}{x}\right)^2-2$
$=196-2$
$=194$
$\therefore\left(x^2+\frac{1}{x^2}\right)=194$
$\left(x^2+\frac{1}{x^2}\right)=\left(x+\frac{1}{x}\right)^2-2 \dots..... (1) $
We first find out $x+\frac{1}{x}$
$x+\frac{1}{x}=(7+4 \sqrt{3})^x+\frac{1}{(7+4 \sqrt{3})}$
$=\frac{(7+4 \sqrt{3})^2+1}{(7+4 \sqrt{3})}$
$=\frac{49+48+56 \sqrt{3}+1}{(7+4 \sqrt{3})}$
$=\frac{98+56 \sqrt{3}}{(7+4 \sqrt{3})}$
$=\frac{14(7+4 \sqrt{3})}{(7+4 \sqrt{3})}$
$=14 $
substitutingin $(1)$
$ \left(x^2+\frac{1}{x}\right)^2=\left(x+\frac{1}{x}\right)^2-2$
$=196-2$
$=194$
$\therefore\left(x^2+\frac{1}{x^2}\right)=194$
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