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?
$\endgroup$ 23 Answers
$\begingroup$$a\sqrt{bc} = a\sqrt{b}\sqrt c$.
It is FALSE that $a \sqrt{bc} = a\sqrt b\times a\sqrt c$. There is only one $a$; not two.
$6\sqrt{24} = 6\sqrt{4\times 6}= 6\sqrt 4 \times \sqrt 6$.
Your calculation $6\sqrt{4\times 6} = (6\sqrt{4})\times (6\sqrt{6})$ is just plain wrong.
$\endgroup$ $\begingroup$$$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.$$
$\endgroup$ $\begingroup$Notice while simplifying the radicals you are multiplying 6,7 twice...
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