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Solving Rational Equations with Like Denominators - Problem 2
There’s lots of different ways to do Math problems as you guys already know. I’m going to show you two different ways to do this problem. I’m going to show you the long way first and then I’m going to save the shortcut way for the end. Okay so the long way.
Most students when they look at this they think oh okay two equal fractions I’m going to cross multiply. Remember cross multiplying means you’re looking at the products of the diagonals and writing those as equal so those guys multiplied being careful with parentheses are going to be equal to my other diagonals multiplied.
Go through and distribute, okay I’m going to combine, write my equation so that it said equal to 0. 2x to the third minus 2x² minus 2x plus 2 equals 0 and then here I have a polynomial with four terms, so the student who chose this method is kind of unhappy, they’re thinking factoring by grouping, they’re seeing this common factor of 2 blah, blah, blah. This method will work, it’s a little bit tricky though. To factor by grouping I’m going to leave you guys with that.
Next I’m going to show you the shortcut way. You can go through and do this and you will get the right answer, but look at the short cut you guys. Both of those fractions have the exact same denominator, so if I were to multiply both sides of the fraction by x-1, that’s essentially eliminating my denominators. What I have is just to solve 2x² equals 2. Oh my gosh that’s like so easy x² equals 1 here is where the trick comes in. x is going to be the square root of 1, but make sure you’re thinking about the positive and negative versions. I think x could equal to 1, or -1.
Okay most students would stop there and they think they have it right, they be thinking like hey I’m an A+ kid because II saw the short cut well they’re right except for this one last thing you need to think about with rational expressions and that’s excluded values.
Remember an excluded value is anything that would make the denominator equal to 0 or make your function undefined. Check it out if I plug in 1 here, that’s bad news. 1 take away 1 is 0 in my denominator that means that this is an excluded value x cannot be equal to 1. Let’s try-1, -1 take away 1 that would work that means I have -2 in my denominator, same thing here which means -1 is my only answer. +1 was the excluded value.