Solve and find the extraneous solution

let's go ahead and solve this rational equation and then see if we can find some extraneous solutions so the first thing that we always want to do when solve a rational equation is get rid of the fractions notice here we have a trinomial binomial binomial it just doesn't look like that much fun does it so the way to get rid of our denominators is to find something that our denominator is going to evenly divide into now it might be obvious this is going to divide into itself right this will divide into itself and this will divide into itself but what we want to do is find the least common denominator or you could call it the least common multiple like what is the multiple of all three of these denominators that they can all evenly divide into so the first thing you want to look out for is do these two binomials give me this because if this times this equals that that means this divides into that and that divides into that right just think about it this way if I had 12 is equal to a 4 * 3 well then I know 12 is divisible by 4 and 12 is divisible by 3 and obviously we know that 12 is going to be divisible by itself so if we were to kind of multiply this and I'm just thinking in my head right that's going to multiply to give me -2 3x * -3 plus a 2X is going to give me negative x 3X * X is a 3X s guess what this actually works out perfectly so and that usually is the case ladies and gentlemen it's not always but it's a great thing for you to always kind of look out for when you are trying to identify the least common denominator least common multiples when you're dealing with pols in your denominator so I'm going to look at this in the factored form and now basically what I need to do is multiply everything by this denominator because this is going to divide into itself leave me with this numerator this is going to divide into that leave me with an x - one and x -1 is going to divide into this leave me with a 3x + 2 so if I was to multiply everything times a 3x + 2 * an x -1 then I can see things are going to battle again these two denominators or denominator and the numerator are going to divide completely out leave me with a 4x^2 - 2 4X when I multiply this times this expression the 3x + 2s are going to divide out leave me with a 3 * x -1 and then over here the x - 1's divide out leave me with a -4 * a 3x + 2 okay now I created a good mess but that's okay so the next thing you need to do is just go ahead and simplify things out I noticed that I have a quadratic term that's my only quadratic term and when you have a quadratic term basically the best way to be able to solve that is to go ahead and see all right let's go ahead and get everything simplified to the quadratic form right ax^2 plus BX plus C and we want to set it equal to zero we're going to want to First Look for factoring but if we can't Factor it then of course we can always rely on the quadratic formula so let's go ahead and simplify and then get everything over to the same side all right and there's a lot going on here but let's go ahead and I can go and switch those up so 4x^2 let's see that's it's going to give me a - 21x okay so now I can get these over to the other side okay so I know a lot of you might want to go oh I'm just going to do Quadra form I don't want to factor that but let's not jump to things so quickly right I know when we have a four sometimes thing gets a little bit difficult right but notice that we have our c is five so there's only so many options to multiply by five right we have five and one and we have just have five and one actually that's our only two options so let's go ahead and start with the 4X and the X so I know if I was to factor this these first two terms could be 4X * X right so I obviously I don't want to put a five here because 4 * 5 would be a 20 it's never going to give me ne9 right and since that's a negative I know my two terms have to both be negative so I don't want to put a neg five here but so what if I put a neg5 here and a negative 1 there let's see if this works we know 4X * X is going to give me a 4x^2 5 * 1 we know it gives us a posi 5 let's go ahead and look at the inner and the outer do these two add up to give me 9x5 * X is a 5x 4X * 1 is 4X 4x + 5x is guess what -9x perfect so now we have a product equal to zero now to go ahead and solve that we can use the zero product property okay so now we can just go ahead and solve to find our two solutions all right and before we just go and say hey we're all done we got to make sure we take these Solutions plug them back into our original equation and make sure that neither of these Solutions make our denominator equal to zero if they do make the denominator equal to zero then they are extraneous now obviously we could always plug them back in and be able to see if our work is correct but if you notice here when I plug a one 1 - 1 is going to be zero so guess what ladies and gentlemen this is extraneous so therefore the only solution is going to be x = 5 - 4 and if you're interested to see me actually plug X5 over4 back into this equation to see if it actually works and we did our math correct then check out the next video I have for you here