Question

If $x=8+3\sqrt{7}$, then find the value of $x^2+\frac{1}{x^2}$

Answer 1

$x=8+3\sqrt{7}$

$x^2=(8+3\sqrt{7}{)}^2$

$x^2=8^2+(3\sqrt{7}{)}^2+2\times 8\times 3\sqrt{7}$

$x^2=64+63+48\sqrt{7}$

$x^2=127+48\sqrt{7}$ .......................(1)

$\frac{1}{x^2}=\frac{1}{127+48\sqrt{7}}$

Rationalising the denominator on the RHS,

$\frac{1}{x^2}=\frac{1}{127+48\sqrt{7}}\times \frac{127-48\sqrt{7}}{127-48\sqrt{7}}$

$\frac{1}{x^2}=\frac{127-48\sqrt{7}}{{127}^2-(48\sqrt{7}{)}^2}$

$\frac{1}{x^2}=\frac{127-48\sqrt{7}}{16129-16128}$

$\frac{1}{x^2}=127-48\sqrt{7}$ .................(2)

Adding (1) and (2)

$x^2+\frac{1}{x^2}=127+48\sqrt{7}+127-48\sqrt{7}$

$x^2+\frac{1}{x^2}=127+127=254$

$x^2+\frac{1}{x^2}=254$