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

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?

• Your first line: $6\times \sqrt {24}=6\times \sqrt 4\times 6 \times \sqrt 6$ is incorrect. – lulu Mar 1 at 16:42
• As a sanity check, you could just evaluate your expression numerically. That's clearly not what's intended but it would certainly clarify which answer was correct. – lulu Mar 1 at 16:43

$$6\sqrt{24} + 7\sqrt{54} - 12\sqrt{6}=6\cdot2\sqrt6+7\cdot3\sqrt6-12\sqrt6=21\sqrt6.$$ I used the following law.

$$a(bc)=(ab)c.$$ For example, $$6\cdot2\sqrt6=(6\cdot2)\sqrt6=12\sqrt6.$$

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

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.

Notice while simplifying the radicals you are multiplying 6,7 twice...