 ###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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# Solving Quadratic Equations in Disguise - Problem 3

Carl Horowitz ###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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We're used to solving quadratic equations where we’re just dealing with ax² plus bx plus c, and we also know how to deal with quadratic equations where we’re dealing with the difference of squares like x² minus 64, something like that. What we want to look at now is this quadratic equation where we have two binomials squared that we’re dealing with. Really what I need you to see is that we’re just dealing with the difference of squares. We have something squared minus something else squared.

First thing students often want to do when we’re looking at a problem like this is students just multiply everything out, combine like terms and then solve it as you would a normal quadratic. I do want to show you a short cut though.

This is just something squared minus something else squared which we know how to factor. Just give you an example, x² minus y² is just equal to x minus y, x plus y. What we really have is just a uglier version of the same thing. This is one thing squared, this is another thing squared, so it’s going to factor exactly the same as this. So just doing that we end up with 2x minus 1 minus 1 plus 3x, times 2x minus 1 plus 1 plus 3x. And then all we have to do and this is equal to zero then all we have to do is simplify each parenthesis and solve it out.

So remember to distribute negative sign through, so we end up with 2x minus 1 minus 1 minus 3x. 2x minus 3x is –x minus 2. And then over here what we end up with is 2x minus 1 plus 1 plus 3x, our 1s cancel just leaving us with 2x plus 3x or just 5x.

What we end up with then if we put these two back together, is 5x times –x minus 2 is equal to zero. So by recognizing we had the difference of squares we were able to just simplify linear statements instead of turning it into a quadratic. Now we just have to solve this, multiplying two things together to equals zero, one or both has to be zero, 5x, that’s easy enough that’s just going to give us x is equal to zero. -x minus 2, it’s a little bit more complicated but hopefully you can see that if we plug in -2, negative -2 is 2 minus 2, that gives us zero. So we end up with zero or negative 2.

Solving a quadratic in disguise, you really just look for patterns in things that you already know and try to make it fit the mode that you can relate to.