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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.$$x^{2}+2 x+10=0$$

$$-1 \pm 3 i$$

Algebra

Chapter 0

Reviewing the Basics

Section 3

Completing the Square

Equations and Inequalities

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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 to solve this quadratic X squared plus two X plus 10 equals zero. And what I like to do is just rewrite this problem with a blank right here. I'm also gonna put a blank on this side. But before I do that, I'm going to subtract 10 over. Cancel that album zero minus 10 is negative. 10. So what goes in that blank? Well, I have to do complete the square, which is taking this middle Number two divided by two is always divide by two and square that number to divide by choose one squared is still one. So I'm going to add one to the left side. But you can't just add one into a problem. You also have to add one to the other side, adding wonderful side so it stays equal. The reason why you do that is that on the left side you have a perfect square china mule that factors have one that add to be too are one and one. There's your short. It's always this number to divide by two. Um, if you don't believe me, foil this out. X plus one times X plus one will give me that piece up there and then on the right side, Negative 10 plus one is negative. Nine. So now what we're ready for is to square root both sides to cancel out that squared. So the left has just X plus one on the right side, whenever you square root, when you're solving, you have a plus or minus. We know the square root of nine is three, but we need an imaginary piece. Just a reminder that there is no way of squaring the number to get a negative. That's why we have to have an imaginary um otherwise you might have like a no real Groot non real solution or something. Statement. But anyway, we're almost done. All we have to do is subtract one over and how we write complex numbers as we put the real part in front of the imaginary part. And that's all we can do for this problem. Yeah,

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