In Exercises 127–130, solve each equation by the method of your c...

Question

In Exercises 127–130, solve each equation by the method of your choice.

1/(x^2 - 3x + 2) = 1/(x + 2) + 5/(x^2 - 4)

Accepted Answer

Welcome back. I'm so glad you're here. We have this equation we get to solve for X by any method we want. So let's see what we've got we've got one over X squared minus eight, X plus 15 Okay, equals one over X minus five minus five Over X squared minus 25. Okay, well let's start off by factoring our denominators and seeing what we've got, factoring this first one here. What times what gives us X squared? That's X times X. What times what? Two factors multiply to equal positive 15 and add up to a negative eight. That's going to be a minus five and a minus three. And I'm going to bring this X minus five down just so we don't forget it. And then over here we recall from previous lessons, this is a difference of squares. So that's going to be an X plus five and an X minus five. Okay, We've got fractions. So one thing we want to note is that we have domain restrictions here and x can't be specific things. What are our different options? What are different factors here? Well, we've got an x minus five and let's set them equal to zero to see what we can't have our exes be um We've got an X minus three and we've got an X plus five. Those are three different factors we've got going on here. So if we add five to both sides. If X equals five, that won't work If we add three to both sides. If X equals three, that won't work and my lines don't there, that's a little better. And if we subtract five from both sides, if X equals negative five that will not work. So something to keep in mind for the end here. And those three factors that we've set, what we want to do is we want to multiply the numerator by those three factors, X minus five, X minus three and X plus five because this will help us get rid of our fractions, get those denominators out of there. So what does that look like? That will be a one over and we'll use our factored forms X minus five times X minus three. And now that's getting multiplied by all three of those factors X minus five times X minus three times X plus five. Okay, there's the first one equals one over. Well that's just x minus five in the denominator. But now all three of those factors in the numerator, X minus five times X minus three times X plus five. And now back to black - over. X plus five times X minus five. And in the numerator will have x minus five times X minus three times X plus five. Okay, now the fun part, let's cancel these factors out. Alright, X minus five. In the numerator and the denominator X minus three in the numerator and the denominator X minus five in the numerator and the denominator X minus five in the numerator And denominator and X plus five in the numerator and the denominator. Now look this is beautiful one times X plus five is just X plus five over here on this left hand side of the equation, one times well I'm just going to bring this down so we don't lose ourselves in our steps. We've got x minus three times X plus five from that first one minus a five times X minus three is all that's left there. Now let's do our distribution. We can foil this out. So that's going to be an X squared for our first are outside will be a plus five X. Our insides negative three times X will be a minus three X. And our last negative three times five is going to be a minus 15 bringing down these other things. So X plus five equals X squared plus five X minus three X minus 15. And distributing that negative five minus five X negative five times negative three is plus 15. And now more fun, more canceling stuff out because look, we've got a plus five X and a minus five X. We've got a minus 15 plus 15. Nice. Okay, let's bring this X over to the right hand side and this five over to the right hand side so that one side will be equal to zero and we can maybe do some factoring or a quadratic equation. And we keep that X with the highest degree. The X squared positive by bringing it all to the right hand side. So we've got zero equals x squared negative three. X minus X is going to be a minus four X. And then just -5. Well, can we factor that? Yeah right. What times what gives us X squared. That'll be X times X. And what times what? Two factors multiply to give us a negative five but add up to give us a negative four. Well that's going to be a minus five plus one. Now we can set each of these two factors equal to zero, X minus five equals zero and X plus one equals zero. We'll bring that five over to the right hand side. So we've got X equals five and we'll bring that one over to the right hand side, subtracting one from both sides and we get an X equals negative one. And yay, that's really exciting. Except hold on. What did we say? We were dealing with this fraction. So we've got domain restrictions and look at this x equals five. That was one of our solutions. That cannot work. We can't have that. Sorry, x equals five. We throw you out because you would give us a domain that just can't exist. You give us a zero in the denominator. So our only solution is X equals negative one. We look at our answer choices and that matches with answer choice B Well done. We'll catch you on the next one

In Exercises 127–130, solve each equation by the method of your choice. 1/(x^2 - 3x + 2) = 1/(x + 2) + 5/(x^2 - 4)

Key Concepts

Rational Functions

Finding Common Denominators

Solving Quadratic Equations

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