Full solution

Q. If

a=\frac{3+\sqrt{5}}{2}

, then find the value of

a^{2}+\frac{1}{a^{2}}

Square a value: Square the value of $a$ . We have $a = \frac{3 + \sqrt{5}}{2}$ . To find $a^2$ , we need to square this expression. $a^2 = \left(\frac{3 + \sqrt{5}}{2}\right)^2$ $a^2 = \frac{(3 + \sqrt{5})^2}{2^2}$ $a^2 = \frac{9 + 6\sqrt{5} + 5}{4}$ $a^2 = \frac{14 + 6\sqrt{5}}{4}$ $a^2 = \frac{7}{2} + \frac{3\sqrt{5}}{2}$
Find reciprocal and square: Find the reciprocal of $a$ and then square it. $\newline$ To find $\frac{1}{a^2}$ , we first find the reciprocal of $a$ , which is $\frac{2}{3 + \sqrt{5}}$ , and then square it. $\newline$ $\frac{1}{a^2} = \left[\frac{2}{3 + \sqrt{5}}\right]^2$ $\newline$ $\frac{1}{a^2} = \frac{4}{(3 + \sqrt{5})^2}$ $\newline$ $\frac{1}{a^2} = \frac{4}{9 + 6\sqrt{5} + 5}$ $\newline$ $\frac{1}{a^2} = \frac{4}{14 + 6\sqrt{5}}$ $\newline$ $\frac{1}{a^2} = \frac{4}{7 + 3\sqrt{5}}$ $\newline$ Now, to rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator, which is $(7 - 3\sqrt{5})$ . $\newline$ $\frac{1}{a^2}$ $0$ $\newline$ $\frac{1}{a^2}$ $1$ $\newline$ $\frac{1}{a^2}$ $2$ $\newline$ $\frac{1}{a^2}$ $3$ $\newline$ $\frac{1}{a^2}$ $4$ $\newline$ $\frac{1}{a^2}$ $5$ $\newline$ $\frac{1}{a^2}$ $6$
Add squared values: Add $a^2$ and $\frac{1}{a^2}$ .
Now we add the expressions we found for $a^2$ and $\frac{1}{a^2}$ .
$a^2 + \frac{1}{a^2} = \left(\frac{7}{2} + \frac{3\sqrt{5}}{2}\right) + \left(7 - 3\sqrt{5}\right)$
To add these, we need a common denominator. Since the second term has an implicit denominator of $1$ , we can multiply it by $\frac{2}{2}$ to get the same denominator as the first term.
$a^2 + \frac{1}{a^2} = \left(\frac{7}{2} + \frac{3\sqrt{5}}{2}\right) + \left(\frac{14}{2} - \frac{6\sqrt{5}}{2}\right)$
$a^2 + \frac{1}{a^2} = \frac{7 + 14}{2} + \frac{3\sqrt{5} - 6\sqrt{5}}{2}$
$a^2 + \frac{1}{a^2} = \frac{21}{2} - \frac{3\sqrt{5}}{2}$