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Overview of the Different Methods of Solving a Quadratic Equation - Problem 2

Teacher/Instructor 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!

Solving a quadratic equation. So I typically say the first step in solving a quadratic equation is to get all of our terms to one side. In this particular problem we already have that taken care of. But this one’s a little bit different because we have a term squared. Whenever we already have something squared the easiest way to get rid of a squared is to square root it, okay.

We could actually FOIL all this out, combine like terms and then either factor, quadratic or complete the square but because something is already squared. it’s easier just to get that by itself. So I want to get this by itself which means I want to add 10 over and then I also need to divide by 5 so I’m going to do that all at once. Add 10, divide by 5 what we’re left with is our squared term equal to 2. We get 2x minus 3² is equal to 2.

Whenever we have something squared in order to get rid of the squared, we want to take the square root of both sides. What that does is cancel the squared over here leaving us with 2x minus 3 and whenever we use the square root as a tool, whenever we put the square root in there, you have to remember your plus or minus. So we have plus or minus square root of 2.

The last thing we want to do is solve for x which means we just have to add 3 and then divide by 2, leaving us with x is equal to plus or minus square root of 2 plus 3 all over 2.

Sometimes teachers will want you to rewrite this as two different things because this is actually two different answers, we could have root 2 plus 3 over 2 or negative root 2 plus 3 over 2 but often times just writing it as plus or minus is perfectly acceptable as well.

So by identifying that we have a squared term we were able to solve this out without having to FOIL it out and just by using our square root property.

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