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In this video, we solve this rational equation, which leads to solving a quadratic equation.
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Alright, we have a rational equation, 3 over x plus 4 over x plus 2 equals 2.
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So, when you have an irrational equation, there are a couple ways to solve it.
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One is to multiply both sides by the least common denominator to eliminate all the fractions.
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I'm going to show you another way to do this, which you might find easier.
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What we could do is rewrite the problem so all of the terms on each side have the same denominator,
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the least common denominator. So, watch what that would look like.
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3 over x and leave little space.
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We've got 4 over x plus 2, leave little space, and I'm going to write 2.
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So, here's the original problem. And I look at this and I think, well, what would the least common denominator?
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Remember 2 would be written as 2 over 1.
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So, the least common denominator is, well, I have an x and an x plus 2.
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So, I'm going to make all of them have the same denominator.
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So, I would have to multiply this first term by x plus 2 over x plus 2.
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This term, I'd have to multiply by x over x.
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And on the right-hand side, I'd have to multiply the bottom by x times x plus 2,
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which means I have to multiply the top also.
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Now, here's the key here. This is an equation.
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So, if you have an equation where both sides really have the same denominator,
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then the numerator should be equal.
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Another way of thinking about this is since they all have the least common denominator,
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if you multiply both sides by the least common denominator,
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it would cancel with every single one of the denominators.
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This becomes a simpler equation, 3 times x plus 2,
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plus 4x equals 2x times x plus 2.
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This only is true if you have an equation, remember?
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If there was no equals 2, you just can't make the denominators disappear.
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It's really because if two numerators are equal, here we are.
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Let's say I have 2 over x equals y over x.
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The denominators are the same.
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Does it make sense that y would have to equal 2?
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That's really what I'm saying here.
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Okay.
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Now, you would get the same equation if you had have taken this original problem,
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and multiplied both sides by x times x plus 2.
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It does take a little bit more space, but you will get exactly the same thing.
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I just wanted to show you another way of looking at the problem, rational equations.
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Okay.
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Now, let's solve this.
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We've got 3x plus 6 plus 4x equals 4x squared,
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I'm sorry, 2x squared, plus 4x.
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By the way, if I would have made a mistake there, later when I checked my answer,
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I would have found out that it was wrong.
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So, remember, if you do make a mistake, that's why I support and always check your answers.
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All right.
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So what do we have here?
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Do you notice you have a plus 4x on both sides?
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So basically, when I subtract 4x from both sides, that's not going to be there anymore.
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What I've got here is a quadratic equation.
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3x plus 6 equals 2x squared.
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So, to solve a quadratic equation, you just said it equal to 0.
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So I'm going to subtract 3x from both sides, and subtract 6.
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And now I need to state the values of A, B, and C.
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A is 2, B is negative 3, C is negative 6, and now we're just going to solve this quadratic equation.
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Okay.
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So, I'm going to do B squared minus 4 AC first, because it just makes less of a mess when I put it into the quadratic formula.
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B squared, well, B is negative 3, so that'll be 9 minus 4 AC.
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And what's A times C? 2 times negative 6?
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That is a negative 12.
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So I have 9 plus 48, or 57.
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Okay.
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Remember, B squared minus 4 AC, or 57, is what's going to go underneath the square root in the quadratic formula.
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So the quadratic formula is negative B, which is, well, B is negative 3, so the opposite that is 3.
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Plus or minus the square root of 57.
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All over 2A. A is 2, so 2A is 4.
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And so I have two solutions.
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3 plus square root of 57, over 4, and 3 minus square root of 57, over 4.
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And actually, this would be really hard to check, so I'm not going to do it on here.
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It's probably easier to do the problem over.
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So those are your solutions, and that's all you have to do.
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Now, I'm going to do that problem again, multiplying by the least common denominator for those of you who may have gotten a little confused when I use this new method.
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All right, so notice the method I use is since an equation, I just made all the denominators the same, so I have a common denominator, and then I made the numerator's equal.
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All right, we're going to do the same thing, the traditional method, multiplying both sides by the least common denominator, which it would be the x times x plus 2.
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All right, so here's the same problem.
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The traditional method is multiply both sides by the least common denominator, which is the LCD is x times x plus 2.
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So I'm going to multiply each term by x times x plus 2, so I can take the 3x, 3 over x, and multiply it.
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Right, and then the next term, it's easier if you write that x plus 2 in parentheses, times x times x plus 2, and then the last, the number on the right hand side also has to be multiplied by the least common denominator.
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And then we see if anything cancels, which of course, anything that was a fraction should cancel if you multiply by the least common denominator, so that x is cancel here, the x plus 2 is cancel here, and there is no fraction on the right, so nothing cancels.
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So I get 3 times x plus 2, 4 times x equals 2x times x plus 2, so I get exactly the same thing as I got up here, there it is.
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3 times x plus 2 plus 4x equals 2x times x plus 2, and here it is again, 3 times x plus 2 plus 4x equals 2x times x plus 2.
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So from this point, everything is the same, you would still use the distributive property, set the equation equal to 0, state the values of a, b, c, plug it in the quadratic formula, and solve, and so this was just verifying that our answer here was correct.
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So that's the answer to this problem. So we started off with a rational equation, and we eliminated the fractions, or two ways I showed you to eliminate the fractions and get this equation right here without fractions.
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And then we ended up with a quadratic equation, use the quadratic formula to solve, because it didn't factor, if it factored, I would try factoring.
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First usually, it doesn't factor, so the quadratic formula seemed the natural way to go here.
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So this is a rational equation that leads you to a quadratic equation, and then you could use the method for solving a quadratic equation.