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Solve each of the quadratics by first completing the square. When the roots are irrational, also give the solutions to the nearest one thousandth.$$x^{2}-3 x+3=0$$

$$\frac{3 \pm \sqrt{3} i}{2}$$

Algebra

Chapter 0

Reviewing the Basics

Section 3

Completing the Square

Equations and Inequalities

Missouri State University

Harvey Mudd College

University of Michigan - Ann Arbor

Lectures

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Solve each of the quadrati…

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In each of the following, …

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we have this quadratic X squared minus three X plus three equals zero, and the way I like to complete the square is just to move the constants to the other side. Sometimes I'll just attracting three over here, so we're looking at negative three. I also like to put in blanks in here because what that helps you do is identify that the number which is found by taking them be value divided by two squared needs to be added to both sides of the equation. And when you square a fraction and multiply the top by itself, So that's positive. Nine negative times Name was positive and two times two is forward. And if I add a number to the left side, I also have to add that same number to the right side. So in completing the square on the left side, we have a perfect square. Try no meaning that it's not. Plus, it's negative and the shortcut. It's always then be value divided by two that goes right there and on the right side. I would change negative three to be negative 12 force. That way I get the same denominator and I can add it to nine force so may have 12 plus nine is three. The denominator stays the same. So now we can square root both sides, and when you square the fraction in your square at the top and bottom. So we have excerpts that should have been a negative. Shouldn't negative right there. That's super important because we can't square root of negative unless we put an eye in there. So we have I wrote three all over to. If you're curious, why we have eyes is the only way to square a number. Whether it's positive, negative or zero, we only get positive support. Zero. So to fix that, we can pull an eye in front to get a complex route. So then your final answer would be to add three halves to the right side. And what's nice is the denominators are the same, so we can just combine them together. And here is your complex solution. Yeah,

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