# 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:

$\sqrt{28}+\sqrt{7}=\sqrt{A}$

• Is it $\sqrt{28} + \sqrt{7} = \sqrt{a}$ and you want to isolate $a$? Hint: $\sqrt{28} = 2 \sqrt{7}$. Now think about adding things and squaring things on both sides of the equals sign. – Moo Dec 23 '15 at 13:46
• @Variable yes I want to isolate a – AugieJavax98 Dec 23 '15 at 13:49

We know that $28 = 4 \cdot 7$ so $\sqrt{28} = 2\sqrt{7}$.

Hence $$\sqrt{A} = 2\sqrt{7} + \sqrt{7} = 3\sqrt{7} \implies A = 9 \cdot 7 = 63.$$

Even if you don't see that $\sqrt{28}=2\sqrt{7}$, you can do algebra as usual: $$A=(\sqrt{28}+\sqrt{7})^2=28+2\sqrt{28}\,\sqrt{7}+7 =35+2\sqrt{28\cdot 7}=35+2\sqrt{196}=35+2\cdot 14=63$$ More generally, if $\sqrt{A}=\sqrt{x}+\sqrt{y}$ you have $$A=x+y+2\sqrt{xy}$$

Here is how it works

$$\sqrt{28}=\sqrt{4 \cdot 7}=\sqrt{4}\sqrt{7}=2\sqrt{7}$$

and hence

$$\sqrt{A} =\sqrt{28}+\sqrt{7}=2\sqrt{7}+\sqrt{7}=3\sqrt{7}$$

$$(\sqrt{A})^2=(3\sqrt{7})^2 \\ A=9 \cdot 7 =63$$