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Solve and check each of the equations.$x(x-2)+2=1$

$x=1$

Algebra

Chapter 1

THE INTEGERS

Section 7

Quadratic Equations with Integral Roots

The Integers

Equations and Inequalities

Polynomials

Missouri State University

Oregon State University

Lectures

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In mathematics, the absolu…

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So you want to solve and check this polynomial this equation. So for someone to distribute this X here so that we can combine like terms to get all numbers on one side So by distribute will have X squared when it's two X and then bring on that place too. And I want us to check one from both sides so we'll have X squared minus two X plus one goes to track the one from both sides because we need to get that equal zero. We're here. So they were gonna factor. Now my factor. And what a separate this middle term here. So anything number cymbals apart. You are one times one when she's blind and we need them to add to negative too. Lookie lous. The backers of one that's just one and one and because we're multiplying neighborhood have to be positive, for they both have to be negative. But in this case, there was need to be negative because we have a negative too sad, too in native one minus one against are making it too. So now we're going to explain this X squared minus X minus X. They bring down that plus one and then we get the first to go to second too. Don't forget to include that negative there. They're never gonna factor out the X. So we have X times X minus one and then here in a factor out of night. Just wine. Well done. Times X minus one. Sit tight, equal to zero. But because we have the same term here in the same term here, we know we'll have X minus one, and then we bring down what's left. And because these two are the same factor, this is really X minus one. All square people zero So simplified this even more we need to solve for X. It's a d. So we'll take the spirit of both sides. I don't have X minus one equals zero. And if we add one of both sides, we get X equals positive one. But now I need to check to make sure this is the right solution. Someone unpleasant entity. Very original grazing here. So we have one time sporting minus two. Oh, close to instead of equal to one. So have blind times. Negative boy in close to this is being a different place too. And that'll give us one. That's correct. Therefore, X equals one is a current solution

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