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Solving Quadratic Equations in Disguise - Problem 1
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We’re going to look at solving an equation where we’re dealing with fractional exponents and I called this kind of problem a quadratic in disguise because when we actually look at it a little bit deeper this actually turns out to be a quadratic equation.

And the way we can do that is that there are two ways of looking at it. You can either just see it right away which hopefully in time you’ll get to or you can make a substitution. And what you want to do is make a substitution for this middle term, the middle x term. So in this case I want to let u equal x to the 1/3. So that becomes 2u. We then need to figure out how to get x to the 2/3 in terms of u, but if we take x to the 1/3 and square it, power to a power we multiply it this turns into x to the 2/3 which is just the same thing as u².

So by letting u equal this middle x term what I’ve done is I’ve turned my sort of weird looking equation into just a standard quadratic. U² plus 2u minus 3 is equal to zero. Hopefully in time the more you get used to it you’ll just be able to jump into factoring this without having to use substitution but for now let’s start by u in there just to make our lives a little bit easier.

We now have a quadratic equation, w know how to solve this. Either factor or quadratic formula, this one factors nicely leaving us with u plus 3 times u minus 1 is equal to zero.

We have a couple ways we can go from here. We can either substitute back in for x, we know that u is equal to x to the 1/3 so we can plug that back in and then solve or we can solve for u and then set that equal to our x term. Either one is perfectly fine. In general I like to solve for my u term.

So I know that u then is equal to -3 or 1. So when we solve for u the question has xs so we have to make sure we substitute in and solve for x. U is equal to -3 or 1, which tells us that x to the 1/3 is equal to -3, or x to the 1/3 is equal to one.

Remember 1/3 is just the cube root, power over root. So in order to solve this out we just have to cube both sides, so the x to the 1/3 is equal to -3, cube that, this will end up x s equal to -27, cube the bottom statement, 1 cubed is just 1 so we end up with x as equal to -27 or 1.

Solving a quadratic in disguise, it’s not that hard just try to make, look for patterns, your middle term squared is equal to your end term then you know you at least have a quadratic in disguise and something to deal with.

What I was talking about is hopefully seeing a pattern is once you’ve seen something like this for a little while, hopefully what you can do is just jump from this term straight to x to the 1/3 plus 3 times x to the 1/3 minus 1 is equal to zero. From here you’ll be able to end up with this equation which you know how to solve.

Solving a quadratic equation in disguise, just making sure you see the pattern that it actually is a quadratic equation.

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