3
$\begingroup$

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:

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

$\endgroup$
  • 3
    $\begingroup$ 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. $\endgroup$ – Moo Dec 23 '15 at 13:46
  • $\begingroup$ @Variable yes I want to isolate a $\endgroup$ – AugieJavax98 Dec 23 '15 at 13:49
5
$\begingroup$

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

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

$\endgroup$
4
$\begingroup$

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 $$ A=x+y+2\sqrt{xy} $$

$\endgroup$
3
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ Thanks everyone for your help! $\endgroup$ – AugieJavax98 Dec 24 '15 at 13:29
  • $\begingroup$ @AugieJavax98: You are welcome. :) Welcome to MSE. You can vote up and check-mark answers. Note the arrows and the check mark next to each answer. :) $\endgroup$ – H. R. Dec 24 '15 at 13:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.