# Struggling to simplify $w^{3/2}\sqrt{32} - w^{3/2}\sqrt{50}$ to $-w\sqrt{2w}$

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

Then:

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

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

So:

$$\sqrt{w^3}4\sqrt{2}$$ - $$\sqrt{w^3}5\sqrt{2}$$ # 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}

• 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? – Doug Fir Jan 6 at 18:00
• Both terms have $\sqrt{w^3}$ as a factor. – KM101 Jan 6 at 18:02
• Well, I used the distributive property $x (a+b)=xa+xb$. You may find this helpful. – 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

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