[Algebra]
Problem: Determine all pairs $(a, b)$ of real numbers that satisfy the following system of equations:

$$
\begin{aligned}
\sqrt{a}+\sqrt{b} & =8 \\
\log _{10} a+\log _{10} b & =2
\end{aligned}
$$

Give your answer(s) as pairs of simplified exact numbers.

Solution: From the first equation, we note that $a \geq 0$ and $b \geq 0$, since the argument of a square root must be non-negative.

From the second equation, we note that $a>0$ and $b>0$, since the argument of a logarithm must be positive.

Combining these restrictions, we see that $a>0$ and $b>0$.

From the equation $\log _{10} a+\log _{10} b=2$, we obtain $\log _{10}(a b)=2$ and so $a b=10^{2}=100$. From the first equation, obtain

$$
\begin{aligned}
(\sqrt{a}+\sqrt{b})^{2} & =8^{2} \\
a+2 \sqrt{a b}+b & =64 \\
a+2 \sqrt{100}+b & =64 \\
a+b & =64-2 \sqrt{100}=44
\end{aligned}
$$

Since $a+b=44$, then $b=44-a$.

Since $a b=100$, then $a(44-a)=100$ or $44 a-a^{2}=100$ and so $0=a^{2}-44 a+100$.

By the quadratic formula,

$$
a=\frac{44 \pm \sqrt{44^{2}-4(1)(100)}}{2 \cdot 1}=\frac{44 \pm \sqrt{1536}}{2}=\frac{44 \pm 16 \sqrt{6}}{2}=22 \pm 8 \sqrt{6}
$$

Since $b=44-a$, then $b=44-(22 \pm 8 \sqrt{6})=22 \mp 8 \sqrt{6}$.

Therefore, $(a, b)=(22+8 \sqrt{6}, 22-8 \sqrt{6})$ or $(a, b)=(22-8 \sqrt{6}, 22+8 \sqrt{6})$.

(We note that $22+8 \sqrt{6}>0$ and $22-8 \sqrt{6}>0$, so the initial restrictions on $a$ and $b$ are satisfied.)

Final answer: $(22+8 \sqrt{6}, 22-8 \sqrt{6})$,$(22-8 \sqrt{6}, 22+8 \sqrt{6})$