19. Let $a, b$ and $c$ be real numbers such that
$$
a=8-b \text { and } c^{2}=a b-16 \text {. }
$$

Find the value of $a+c$.
19. Ans: 4
$$
\begin{aligned}
& c^{2}=\boldsymbol{a}-16=(8-b) b-16=-b^{2}+8 b-16 \\
\Longrightarrow & c^{2}+b^{2}-8 b+16=0 \Longrightarrow c^{2}+(b-4)^{2}=0 \Longrightarrow c=0 \text { and } b=4 .
\end{aligned}
$$

Since $b=4$, it follows that $a=8-4=4$. Thus, $a+c=4+0=4$. In this solution, we have used the fact that the sum of two non-negative numbers is zero only when both numbers are zero.