Solve for the roots of the following equation.

x^2/3 + x^1/3 + 1 = 0

Is there some way to express this as a quadratic equation?

Create a new variable, u, and set it equal to x^1/3.

u² + u + 1 = 0

Hey hey hey, this new equation is quadratic. It isn't factorable, though, so we use the quadratic formula to find the roots.

 and 

These are the roots for the new, u-based equation. Since we already said that u = x^1/3, solving for x gives us x = u³.

 and 

We were right; this is a super nasty problem. Let's go ahead and check our results, though.

Luckily, most of the exponents cancel each other out.

One down.

And the other. That means our answers are  and . It's really satisfying to get a long problem like this right after all this work. It gives us a warm, fuzzy feeling all over.

Solve for the roots of the following equation.

9x⁴ – 4 = 0

Start off by recognizing that this is a quadratic equation in disguise.

9u² – 4 = 0

Let u = x².

 and 

We can isolate u² because we don't have any terms with just u.

 and 

 and 

We should now check that the answers are correct. Like washing your hands after using the bathroom, it's just the right thing to do.

The only x term has an even exponent, so we can check the positive and negative values at the same time.

Solve for the roots of the following equation.

We can't really do anything until we kick that denominator to the curb. Use the LCD to multiply it outta there.

1 + 2x(6 + x) + 4(6 + x) = 0

1 + 12x + 2x² + 24 + 4x = 0

2x² + 16x + 25 = 0

From here, treat this like any other quadratic equation. We suggest using the quadratic formula this time.