Algebra

By Vieta

p + q  =   -5/1 =   - 5

pq  =  3/1   =   3

Note  that

1/p  +  1/q  =     (p + q)  / (pq)  =       -5 / 3

The roots of $x^2 + 5x + 3 = 0$ are p and q.
Find $\dfrac1p + \dfrac1q$.

$\text{Set $x=\dfrac{1}{x}$ in $x^2 + 5x + 3$ }$

$\begin{array}{|rcll|} \hline x^2 + 5x + 3 &=& 0 \\\\ \Rightarrow \left(\dfrac{1}{x}\right)^2 + 5*\dfrac{1}{x} + 3 &=& 0 \\\\ \dfrac{1}{x^2} + \dfrac{5}{x} + 3 &=& 0 \quad | \quad * x^2 \\\\ 1 + 5x + 3x^2 &=& 0 \\\\ 3x^2+5x+1 &=& 0 \quad | \quad : 3 \\\\ x^2+\dfrac{5}{3}x +\dfrac13 &=& 0 \\ \hline \text{By Vieta:}~\dfrac1p + \dfrac1q &=& -\dfrac{5}{3} \\\\ \text{By Vieta:}~\dfrac1p * \dfrac1q &=& \dfrac{1}{3} \\ \hline \end{array}$