Here’s a crazy looking radical problem:

For which of the following values of *w* does $$sqrt[4]{16w^3x^frac9w}=(2)(3^frac34)(x^frac34)$$

A) 2

B) 3

C) 4

D) 6

**Strategy****:**

**Separate terms underneath a radical.****Consider writing the radicals as fractional exponents then using the exponent rules.****Always check for extraneous solutions!**

__Radical Rules__

- This property allows you to pull numbers out from the square root sign: $$sqrt{xy}=sqrt x;times;sqrt y\sqrt{32}=sqrt{2times16}=sqrt2;times;sqrt{16}=4sqrt2$$
- $$sqrt{frac xy}=frac{sqrt x}{sqrt y}$$
- If we have different roots, we have to first convert to exponents to combine. $$sqrt8;times;sqrt[3]8;neq;sqrt[4]8$$
- Instead, we must do $$sqrt8;times;sqrt[3]8=8^frac12;times;8^frac13=8^frac56=sqrt[6]{8^5}$$

Now, back to our crazy example:

This is the kind of problem the SAT wants you to look at and freak out about because it looks so complex. But as always, getting to the answer isn’t so hard if you know your rules!

We see on the right side we have fractional exponents. This is a big clue that we need to convert the left side. We start by getting rid of our radical.

$$(16w^3x^frac9w)^frac14$$

Remember that when we have an exponent raised to an exponent, we multiply. Don’t forget about the coefficient!

$$16^frac14{;w}^frac34;x^frac9{4w}=(2)(3^frac34)(x^frac34)$$

We want the value of *w*. If we look at our middle terms, we can now easily find *w* has to equal 3. We only had to do two steps to find this!

__Extraneous Solutions__

One more note on radicals. Anytime you start with a radical in an equation and square both sides to get rid of it, you risk introducing what is known as an **extraneous solution**. This is a false answer that won’t work in the original equation. For example:

$$sqrt{x-a}=x-4$$

If *a* = 2, what is the solution set of the equation above?

A) {3, 6}

B) {2}

C) {3}

D) {6}

The first thing we need to do is plug in our constant.

$$sqrt{x-2}=x-4$$

Then, a quick look at our answers tells us there are only 3 possibilities (2, 3, and 6). This equation is simple enough that you should just plug them in and solve three times.

However, if you forgot to do that, or just like a little extra pain on your SAT, you could square both sides to get

$$x-2=(x-4)(x-4)\x-2=x^2-8x+16\0=x^2-9x+18\0=(x-3)(x-6)$$

So 3,6 is our solution set. But you already see where this one is going by the section we’re in. If we plug 3 back into our original equation, we get

$$sqrt{3-2;}=3-4\sqrt1=-1\1=-1$$

Clearly false! Always check for extraneous solutions when you square to clear away radicals.

We can’t stress this enough: test your solutions when dealing with radicals or variables in denominators. Almost always, one of the solutions is extraneous (doesn’t work).

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