MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
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MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
In order to solve this equation I’m going to use completing the square and the first thing I want to do is make sure that my leading coefficient or my a value is 1. It’s not, right now is 3. So by absolute first step is going to be you divide everything up here by 3. Keep in mind 0 divided by 3 is still 0. But what I’m going to have is p² take away 4p take away 5 is equal 0. This is equivalent to my original trinomial only now I got rid of all those 3 factors.
Okay so to complete the square after you have your leading coefficient equal to 1, the next thing you do is you get all your p values together and then you get your constant on the other side of the equals. So I added 5 to both sides. One thing I like to do when I'm completing the square is write some little blanks here because I’m going to be adding things to both sides of the equation.
Here’s what I’m going to do to complete the square, take half of this value square it and write my result here. So I’m going to half of -4 which is -2 square it, that’s where I get +4 and add that to both sides of the equation. Now what I have is p take away 2² is equal to 9. Let me back up though and take you through that one more time what I did.
I took half of this -4 term that was -2, then I did that value squared and added that to both sides. Because in order to turn this into a perfect square trinomial, I had to add 4. But you can’t just go on adding 4 whenever you want to, you have to add 4 to both sides of the equation in order to keep that equivalent. That’s how I got this 9.
Okay if you understand that you're in good shape. From now on I’m just going to solve for p by taking the square root of both sides, p minus 2 is equal to the square root of 9 which is 3. and then I’m going to go through and make sure I use the positive value and the negative value of the square root of 9. I’m going to have two different answers. I’ll have to solve p minus 2 equals 3 and also p minus 2 equals -3. So I’ll get the answer p equals -1 that’s one of my possible solutions the other answer is p equals 5.
All right, I think I solved those equations what I want to do is go back and plug in these values to check, so let’s do that. I’m going to kind of do it in my head. Let’s say p was equal 5, I would have 3 times 25 which is 75, take away 12 times 5, take away 15, good that does indeed give me the answer 0.
Let’s check the answer p equals -1. -1² is +1 multiply by 3 is 3, take away -12 times -1 which is going to give me +12 minus 15 equals 0. Good, that told me that both of my answers were correct.
So one last time I’m going to talk you through the tricky part, the first thing I had to do is make sure my leading coefficient was one by dividing everything by 3, then I took half of this b value, squared it and add the result to both sides. From there it's just solving by taking square roots.
So these problems are pretty straight forward in this situation I got whole number answers which tells me I could have factored to begin with but let’s just say I didn’t know that. Completing the squares is a good skill to have especially when you get into the more difficult problems or you have big decimals as your solutions.
Unit
Quadratic Equations and Functions