Question

State the following statement is True or False

If $\alpha =\frac{3+\sqrt{5}}{2}$, then the value of ${\alpha }^2+\frac{1}{{\alpha }^2}$ is $7$.

• A. True
• B. False

Answer 1

The correct option is A True

$\alpha =\frac{3+\sqrt{5}}{2}$
$\frac{1}{\alpha }=\frac{2}{3+\sqrt{5}}$
$=\frac{2(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}$
$=\frac{2(3-\sqrt{5})}{3^2-{\left(\sqrt{5}\right)}^2}$
$=\frac{2(3-\sqrt{5})}{9-5}$
$=\frac{2(3-\sqrt{5})}{4}$
$=\frac{3-\sqrt{5}}{2}$
Now,
${(\alpha +\frac{1}{\alpha })}^2={\alpha }^2+\frac{1}{{\alpha }^2}+2\times \alpha \times \frac{1}{\alpha }$
$=>{\alpha }^2+\frac{1}{{\alpha }^2}={(\alpha +\frac{1}{\alpha })}^2-2$
Now, putting the values of $\alpha$ and $\frac{1}{\alpha }$, we get,
${\alpha }^2+\frac{1}{{\alpha }^2}={(\frac{3+\sqrt{5}}{2}+\frac{3-\sqrt{5}}{2})}^2-2$
$={\left(\frac{3+\sqrt{5}+3-\sqrt{5}}{2}\right)}^2-2$
$={\left(\frac{6}{2}\right)}^2-2$
$={\left(3\right)}^2-2$
$=9-2$
$=7$