$6\sqrt{24} + 7\sqrt{54} – 12\sqrt{6}$ = $21\sqrt{6}$ but I get $207\sqrt{6}$

algebra-precalculusradicals

I'm asked to simplify $6\sqrt{24} + 7\sqrt{54} – 12\sqrt{6}$

The provided solution is $21\sqrt{6}$ but I arrive at a different amount.

Here is my working, trying to understand where I went wrong:

First expression:
$6\sqrt{24}$ = $6\sqrt{4}$ * $6\sqrt{6}$ = $6*2*6\sqrt{6}$ = $72\sqrt{6}$

Second expression:
$7\sqrt{54}$ = $7\sqrt{9} * 7\sqrt{6}$ = $147\sqrt{6}$

Third expression is already the remaining common expression $12\sqrt{6}$.

So: $147\sqrt{6} + 72\sqrt{6} – 12\sqrt{6}$ = $207\sqrt{6}$

Where did I go wrong?

Best Answer

$a\sqrt{bc} = a\sqrt{b}\sqrt c$.

It is FALSE that $a \sqrt{bc} = a\sqrt b\times a\sqrt c$. There is only one $a$; not two.

$6\sqrt{24} = 6\sqrt{4\times 6}= 6\sqrt 4 \times \sqrt 6$.

Your calculation $6\sqrt{4\times 6} = (6\sqrt{4})\times (6\sqrt{6})$ is just plain wrong.