Solving Quadratic Equations by Factoring @MathTeacherGon

hi guys it's mithy kergoin in our today's video we will talk about solving quadratic equations by factoring so we have here this example and later on we will talk about this problem because a lot of you are requesting for gridding topics in mathematics so this is the first topic that you will encounter in solving quadratic equations when you are in greed in mathematics so without further ado let's do this topic so before we start answering this equation or finding the solution of this equation let's try first this first two examples so by the way guys when we are doing solving quadratic equations what we need to do or what is the purpose of that is that we will try we are trying to find the values of x or what are the possible values of x to make this whole equation be equal to zero and one of the process or one of the ways on how to solve quadratic equation is by factoring so let's start with item number one and number two in number and number two we will use greatest common factoring so as you can see we have x squared minus 5 x is equal to 0. so we need to think of a greatest common factor from both x squared and negative 5 x and as you can see when we talk about the coefficients variable we have the common factor which is the variable x and after that to get the other factor what you need to do is you will divide the first term x squared by x again x squared it goes like this x squared over x so x squared over x or x squared divided by x is simply x that is the first term of the second factor and to get the next term of this you will divide negative 5 you will divide negative 5 x by your greatest common factor which is x and negative five x divided by x is negative five so what we have now is what we have here the factors are x times x minus five then equal to zero so what is the next step after factoring so you will equate both factors by zero so you will equate this first factor by zero and that is x is equal to zero actually this one is already considered as an answer so we will name this as x sub one so for the next factor we have x minus five is equal to zero so what we need to do is to manipulate this one since this one is negative five we will add five both sides of the equation so what we have now is this will become zero we have x is equal to zero plus five which is equal to five other content creators what we what they do is that this negative five you they will transpose this term to make it positive five so parallel so what we have here are x sub 1 is equal to 0 and x sub 2 is equal to 5. these are the roots or the solutions of the quadratic equations x squared minus five x is equal to zero so let's continue we have here negative four oh sorry four x squared minus six x is equal to zero so as you can see for from the coefficients we have four and negative six so what we have here the factors are for them from the coefficient we have positive two and as for the variable x squared and x the other is x so our greatest common factor right now is 2x so how to get the second factor we will divide 4x squared by 2x and that will be 2x next negative 6x divided by 2x that is equal to negative 3 and then equate by 0. so if you want to check whether your factors are correct you can reverse the process you can multiply it one by one you can multiply 2x by 2x and that will be 4x squared we have 2x times negative 3 and that is negative 6x so let's continue the first factor is 2x equate by zero then solve for x divide by two and divide by two cancel cancel your x now is zero divided by two is still zero so this is now our x sub one so let's continue for the second factor we have 2x minus 3 is equal to 0. so here instead of adding 3 i will just transpose this to make it easier so we have now 2x is equal to remember when you're transposing a term it will change you need to change the the sign from negative 3 it will become positive 3 okay so what we have now is 2x to be divided by 2 divided by 2 cancel cancel your x is equal to 3 over 2 and this will serve as x sub 2. so now what are the roots the roots of the equation or the quadratic equation is zero and three over two so after item number one and number two let's continue doing item number three and number four for item number three number four we will incorporate um difference of two squares from item number one and number two say you know why not and done we use greatest common factoring this one the maritime pattern and pattern at incanto you have x squared minus y squared when you factor this 2 that is equal to x plus y times x minus y so let's try to get the factors of this you need to get the square root of x squared that is x so prepare time two parenthesis x and x okay to get the second factor or the second term the square root of 49 is seven and seven now follow the pattern this one is positive or plus and this one is subtraction so sir can we interchange the two factors yes is it possible it's possible to have x minus seven times x plus seven and then equal to zero then since we have two factors we can equate both factors by zero that is x plus seven is equal to zero and the other is x minus seven is equal to zero so what's next is we will transpose this to the other side from positive it will become negative so that is x sub one is equal to negative 7 this is the first solution or root of x squared minus 49 let's continue transpose this to the other side from negative it will become positive so that is x sub 2 is equal to positive 7 and these are the roots of x squared minus 49 is equal to 0. we have negative 7 and positive 7. let's move on with number 4. we have three x squared minus seventy-five so as you can see uh we cannot use directly the difference of two squares pero perimenopausal factor as you can see we have three and seventy-five so factor on an atom that is three times three x squared divided by three is x squared negative seventy five divided by three that is minus twenty is equal to zero next as you can see this pattern is the same as this pattern so we can factor out x squared minus 25 like this we have three times prepare two parenthesis and then get the square root of this that is x and x the square root of 25 is five and five so okay badapa plus minus is equal to zero now we have three factors we have three x plus five and x minus five ang equates assume x plus five in x minus five so what we have here is x plus five is equal to zero x minus five is equal to zero transpose this to the other side we have x is equal to negative five as for x sub one this is your first root or solution transpose this is the other side that is x is equal to positive 5 and this will be your x sub 2. and these are the roots of 3x squared minus 75 is equal to zero now let's continue with number five and number six this one is a trinomial so let me remove this paper now as you can see we cannot use greatest common factoring and difference of two squares what we have here is a trinomial where we need to factor out this one but in different pattern okay so let's try whenever i'm encountering this kind of quadratic equations i'm preparing two different parentheses then equal to zero since our first term is x squared or it has the coefficient of one i automatically will place x and x here so my main problem here are the second terms of the two factors so how i will go into how i'm going to go into and to find the second terms of the two factors i will consider first negative 15. so we have negative 15. i will think of factors of negative 15 that will make or that has the sum of negative 2. again i will find the factors of negative 15 that has the sum of negative 2. so what are the possible factors at time of factors net n in negative 15 um the nearest factors are negative five times three and positive five times negative three as you can see the two set of factors are also equal to negative fifteen but what we need to think is that ancestor negative two so again when we try negative five plus three that is negative 2. so definitely these are the possible second terms so what about 5 times negative 3 try to add 5 and negative 3. that will give you positive 2 and what we're trying to figure out is how are we going to get negative two okay so second terms are negative five and plus three so in this case we can continue solving the equations first factor x minus five is equal to zero and then the other is x plus three is equal to zero transpose this to the other side that is x is equal to positive five this is your x sub one next x plus 3 transpose positive 3 to the other side that will become negative that is x is equal to negative 3 that will serve as your x sub 2. and right now this are the solutions or the roots of x squared minus 2x minus 15 is equal to 0. now for number 6 this will be your assignment what are the factors or i mean what are the solutions of x squared plus five x plus six is equal to zero comment down below now let's move on with our first example in our first example we have here x squared plus 3x minus 28. so let's figure out what are the factors first of negative 28 but before that we will prepare two sets of parentheses okay then equal to zero so continuing we have negative 28. for negative 28 our target here is to find the factors of negative 28 that will give us the sum of positive three so when we talk about the possible factors of negative 28 the nearest pairs are negative seven and times four the other pair is positive seven times negative four so again don't get me wrong because 1 times 28 1 times negative 28 or negative 1 times 28 are still possible factors of negative 28 but what we're trying to figure out here is that the factors must give us the positive three so it is try negative seven plus four that is negative three so it's not possible so seven plus negative four that is positive three so what we have now these are the possible factors or possible factors of negative 28 so get the square root of x squared that is x and x next we have plus seven then minus four so since we already have the factors we will equate each factor by zero we have x plus seven is equal to zero and the other factor is x minus four is equal to zero get this one transpose to the other side of the equation that is x is equal to negative seven so this is now your x sub 1. for your second factor transpose negative 4 to the other side this will become positive so x sub 2 is equal to positive 4. so what are these are the possible roots or solutions of the first quadratic equations in a given problem so guys if you're new to my channel don't forget to like and subscribe and also you can follow me satin facebook page across the teacher if you're new to my channel don't forget to like and subscribe but i hit nothing above button for you to be updated saturday latest uploads again it's mitigar gone bye

Channel: MATH TEACHER GON

Views: 561,447

Keywords: quadratic, quadratic equations, solving equations, grade 9, grade 9 math, solving equation by factoring, factoring, solving quadratic equations, math teacher gon, completing the square, how to solve quadratic equations

Id: Vt1bxQ1yXzs

Length: 16min 11sec (971 seconds)

Published: Mon Aug 15 2022