I'm asked to simplify $w^{3/2}\sqrt{32} - w^{3/2}\sqrt{50}$ and am provided with the solution: $-w\sqrt{2w}$

I arrived at $9\sqrt{2}$ but I think I'm confused in understanding communitive rule here.

Here is my working:

$w^{3/2}\sqrt{32} - w^{3/2}\sqrt{50}$ = $\sqrt{w^3}\sqrt{32}$ - $\sqrt{w^3}\sqrt{50}$ # is this correct approach? I made the radical exponent a radical


$\sqrt{32}$ = $\sqrt{4}$ * $\sqrt{4}$ * $\sqrt{2}$ = $2 * 2 * \sqrt{2}$ = $4\sqrt{2}$

$\sqrt{50}$ = $\sqrt{2}$ * $\sqrt{25}$ = $5\sqrt{2}$


$\sqrt{w^3}$$4\sqrt{2}$ - $\sqrt{w^3}5\sqrt{2}$ # should the expressions on either side of the minus sign be considered a single factor? i.e. could I also write as ($\sqrt{w^3}$$4\sqrt{2}$) - ($\sqrt{w^3}5\sqrt{2}$) )?

Then I'm less sure about where to go next. Since I have a positive $\sqrt{w^3}$ and a negative $\sqrt{w^3}$ I cancelled those out and was thus left with $9\sqrt{2}$.

More generally I was not sure of how to approach this and could not fin a justification for taking the path that I did.

How can I arrive at $-w\sqrt{2w}$ per the text book's solution?


Your approach is absolutely right. But note that $$ \begin{align} \sqrt{w^3}4\sqrt{2}- \sqrt{w^3}5\sqrt{2}&=\sqrt{w^3}(4\sqrt{2}-5\sqrt{2})\\ &=-\sqrt{w^3}\sqrt{2}=-w\sqrt{2w}. \end{align}$$

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    $\begingroup$ Thanks for the answer, I follow and understand your solution. How is it that you knew to factor $\sqrt{w^3}$? After I had rewritten $\sqrt{32}$ as $4\sqrt{2}$ and then $\sqrt{50}$ as $5\sqrt{2}$ I was stumped about next steps. Is this just a practice an intuition thing or is there a prescribed set of rules and order of operations that I'm missing? $\endgroup$ – Doug Fir Jan 6 at 18:00
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    $\begingroup$ Both terms have $\sqrt{w^3}$ as a factor. $\endgroup$ – KM101 Jan 6 at 18:02
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    $\begingroup$ Well, I used the distributive property $x (a+b)=xa+xb $. You may find this helpful. $\endgroup$ – Thomas Shelby Jan 6 at 18:11

You are on the right track to simplify $ w^\frac{3}{2} \sqrt{32} - w^\frac{3}{2} \sqrt{50} $ to as far as

$$\mathrm{(1)} \qquad \sqrt{w^3} 4 \sqrt{2} - \sqrt{w^3} 5 \sqrt{2} $$

I would make expression $ (1) $ neater and rewrite as

$$\mathrm{(2)} \qquad 4 \sqrt{w^3} \sqrt{2} - 5 \sqrt{w^3} \sqrt{2} $$

These two terms are alike, and combining the two yields

$$\mathrm{(3)} \qquad -\sqrt{w^3} \sqrt{2} $$

According to one of the properties of radicals, $ \sqrt{a} \sqrt{b} = \sqrt{ab} $. Using that property and rearranging,

$$\mathrm{(4)} \qquad -\sqrt{2w^3} $$

Now $ w^3 $ can be expressed as a product involving a perfect square (i.e. $ w^2 $) and a non-perfect square ($ w $). Expression $ (4) $ becomes

$$\mathrm{(5)} \qquad -\sqrt{2w w^2} $$

Simplify to get the desired result

$$ -w \sqrt{2w} $$


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