Question

If
$a {}^{2} - 5a - 3 = 0$
Then find,
$a {}^{2} + \frac{1}{a {}^{2} }$
Please answer fast. I will mark BRAINLIEST. Please it's very urgent. ​

Answer 1

$\begin{aligned}&3a^2-5a-3&=&\ \ 0\\ \\ \Longrightarrow\ \ &3a^2&=&\ \ 5a+3\\ \\ \Longrightarrow\ \ &a^2&=&\ \ \dfrac{5a+3}{3}\\ \\ \Longrightarrow\ \ &\dfrac{1}{a^2}&=&\ \ \dfrac{3}{5a+3}\end{aligned}$

$\begin{aligned}&a^2+\dfrac{1}{a^2}\\ \\ \Longrightarrow\ \ &\dfrac{5a+3}{3}+\dfrac{3}{5a+3}\\ \\ \Longrightarrow\ \ &\dfrac{(5a+3)^2+3^2}{3(5a+3)}\\ \\ \Longrightarrow\ \ &\dfrac{25a^2+30a+18}{15a+9}\\ \\ \Longrightarrow\ \ &\dfrac{25a^2}{15a+9}+\dfrac{30a+18}{15a+9}\\ \\ \Longrightarrow\ \ &\dfrac{25\left(\dfrac{5a+3}{3}\right)}{15a+9}+2\\ \\ \Longrightarrow\ \ &\dfrac{\frac{25(5a+3)}{3}}{15a+9}+2\end{aligned}$

$\begin{aligned}\Longrightarrow\ \ &\dfrac{25(5a+3)}{3}\div (15a+9)+2\\ \\ \Longrightarrow\ \ &\dfrac{25(5a+3)}{3}\times \dfrac{1}{(15a+9)}+2\\ \\ \Longrightarrow\ \ &\dfrac{25}{3}\times \dfrac{1}{3}+2\\ \\ \Longrightarrow\ \ &\dfrac{25}{9}+2\\ \\ \Longrightarrow\ \ &2\dfrac{7}{9}+2\\ \\ \Longrightarrow\ \ &\Large \dfrac{43}{9}=4\dfrac{7}{9}=4.\overline{7}\end{aligned}$

Hence the answer is 4 and 7/9, which equals 4.777...