Find $A $ in $\sqrt{28}+\sqrt{7}=\sqrt{A}$ I got this problem from an old test paper of mine. I tried working on it today but I couldn't solve it. I tried solving it using logarithms but I didn't get an answer. The problem asks to solve for $A$.Can you guys help me solve this problem?
Here's the problem:
$\sqrt{28}+\sqrt{7}=\sqrt{A}$

Here is how it works
$$\sqrt{28}=\sqrt{4 \cdot 7}=\sqrt{4}\sqrt{7}=2\sqrt{7}$$
and hence
$$\sqrt{A} =\sqrt{28}+\sqrt{7}=2\sqrt{7}+\sqrt{7}=3\sqrt{7}$$
which leads to
$$(\sqrt{A})^2=(3\sqrt{7})^2 \\
A=9 \cdot 7 =63$$
