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
The first step when I’m going to solve something by completing the square is to make sure my leading coefficient is 1. It’s not 1 here it's 4 so what I’m going to do is go through and divide all of these terms by 4 so that my leading coefficient is 1.
Now I’ll have x² take away x plus 9/4, fraction equals 0. Okay continuing with the completing the square process I want the x terms alone on the one side of the equals and I want the constant term on the other side of the equals. Then I’m going to take half of my b value, half of -1 is -1/2, I’m going to square it and then add that value to both sides of the equals. Well one half or -1/2 times -1/2 is +1/4 so there it goes +1/4, +1/4.
So this side represents the perfect square trinomial, here it is written in its factor form and it's equal to -9/4 plus 1/4 which is -8/4. You could probably in your head reduce that to -2. Okay so I’m on my way. next thing I want to do to get x all by itself is take the square root of both sides. So I’ll have x minus 1/2 is equal to the positive and negative square root of -2.
Do you guys see why this is a funny problem? Check it out, I can’t have the square root of a negative number. Well I can but it’s not a real solution. Square root of a negative number tells me I’m done I don’t have to do any more math. No real solutions, I write up those words that’s the answer to my problem. There’s no x value not even a funny long super complicated decimal that when I plug it back here I get the answer 0.
The way I could verify that this is right because I’m kind of nervous no solutions, that’s kind of weird the way I can verify this is by looking at the discriminant of this problem. Remember the discriminant is b² minus 4ac and if my discriminant is negative, then I could confirm that indeed there are no real solutions to this problem.
It’s kind of weird but sometimes in Math you guys you do have the answer, no real solutions just got to go with it.
Unit
Quadratic Equations and Functions