Mastering Quadratic Equations: Factoring and Quadratic Formula Explained

in this lesson we're going to talk about solving quadratic equations by factoring so let's start with this example x

squared minus 49 is equal to zero you can use the difference of perfect squares technique for this one

the square root of x squared is x the square root of 49 is seven so it's going to be x plus seven

and x minus seven now you need to set each factor equal to zero at this point and then you could

find the value of x so we have x plus seven is equal to zero and x minus seven is equal to zero

the reason why we can do that is because if one of these terms is equal to zero then everything is zero zero times

anything is zero so x is equal to negative seven

and in the other equation if we add seven to both sides we could see that x is equal to positive 7.

let's try another example let's say if we have 3x squared minus 75

is equal to zero what is the value of x 3 and 75

are not perfect squares so we don't want to use the difference of perfect squares technique yet however

we can take out the gcf the greatest common factor which is three three x squared divided

by three is x squared negative seventy-five divided by three

is negative twenty-five now we can use the difference of perfect squares technique

to factor x squared minus 25. the square root of x squared is x the square root of 25 is 5.

so it's going to be x plus 5 and x minus 5. so if we set x plus five equal to zero

we can clearly see that x will be equal to negative five and if we set x minus five equal to zero

x is equal to plus five and so that's it for that one now what about this one let's say if we

have 9x squared minus 64 is equal to zero

well first we can use the difference of perfect squares technique we can square root 9

and we can square root 64. the square root of 9 is 3. the square root of x squared is x

the square root of 64 is 8. so it's going to be 3x plus 8 3x minus 8.

so if we set 3x plus 8 equal to 0 then we can see that 3x is equal to negative 8 which means x is equal to

negative eight over three now if we set three x minus eight equal to zero and solve for x

x is gonna be positive eight over three using the same steps now what if we have a trinomial

x squared minus 2x minus 15. and the leading coefficient is one how can we factor this expression

all you need to do is find two numbers that multiply to negative 15 but that adds to negative two

numbers that multiply to fifteen are five and three so we have positive five and negative

three or negative five and three five plus negative three adds up to positive two but negative 5 plus 3 adds

up to negative 2. so this is what we want to use it turns out that

to factor it it's simply going to be x minus 5 plus

x plus 3. so if we set x minus five equal to zero x will be equal to five

and if we set x plus three equal to zero x will be equal to negative three let's try another one like that

let's say if we have x squared plus 3x minus 28

so what two numbers multiply to negative 28 but add to three go ahead and try it

so if we divide 28 by 1 we'll get negative 28 if we divide negative 28 by 2

negative 14 3 doesn't go into it if we divide it by 4 we'll get negative 7. 4 and negative 7 differs by three if we

add them it's negative three so we need to change the sign so it's going to be x minus four times x

plus seven which means that x is equal to positive four

and negative seven here's another problem so how can we factor this trinomial when

the leading coefficient is not one so what we need to do in this problem we need to multiply eight and negative

fifteen eight times negative 15 is negative 120.

now what two numbers multiply to negative 120 but add to two if you're not sure make a list

let's start with one we have one in 120 two and sixty three and forty

four and thirty five and twenty four six and twenty

eight and fifteen now 10 and 12 seem promising 10 and negative 12 differ by negative 2

but positive 12 and negative 10 adds up to positive 2. so what we're going to do in this

problem is we're going to replace 2x with 12x and negative 10x and then factor by grouping

in the first two terms let's take out the gcf which is going to be 4x

8x squared divided by 4x is 2x and 12x divided by 4x

is 3. and the last two terms take out the greatest common factor in

this case negative 5. negative 10x divided by negative 5 is 2x

negative 15 divided by negative 5 that's plus 3. now if you get two common terms

that means you're on the right track you can write it once in

a parenthesis in the next line now the stuff on the outside 4x and negative 5 that's going to go in the

second parentheses so that's what we have now let's set two x plus three equal to

zero and 4x minus 5 equal to 0. so in the first equation let's subtract

3 from both sides so 2x is equal to negative 3. and then let's divide by 2.

so the first answer x is equal to negative three over two now let's find the other answer

so let's add five to both sides so we can see that four x is equal to five

and then let's divide both sides by four so x is equal to five over four

and that's it for this problem now let's get some of the answers to the quadratic equations that we had

in the last lesson so for this particular problem when we factor it we got a solution of 5

and negative 3 in less than 10.2 but now let's use the quadratic equation

to get those same answers so x is equal to negative b plus or minus the square root of b

squared minus 4ac divided by 2a that's the quadratic formula

and you need the quadratic equation in standard form so we can see that a is equal to one

b is the number in front of x b is negative two

and c is negative fifteen so let's replace b with negative two

b squared or negative two squared negative two times negative two is four a is one and c is negative fifteen

divided by two a or two times one which is two negative times negative two is positive

two and then we have four negative four times negative fifteen that's positive

sixty and sixty plus four is sixty-four now the square root of sixty-four is

eight so we have two plus or minus eight divided by two

two plus eight is ten ten divided by two is five that gives us the first answer the next one is two

minus eight divided by two two minus eight is negative six negative six divided by two is negative three

which gives us the second answer so you can solve a quadratic equation by factoring

or by using the quadratic formula now let's try another example eight x squared plus two x

minus fifteen use the quadratic equation to find the values of x

so we can see that a is equal to eight b is the number in front of x that's two c is negative 15.

so using the quadratic formula x equals negative b plus or minus the square root

of b squared minus 4ac divided by 2a so b is 2

which means b squared that's going to be positive 4 minus 4 times a a is 8 c is negative 15

divided by 2 a or 2 times 8 which is 16. so this is negative 2 plus or minus square root

4. now negative 4 times

negative 15 is positive 60 60 times 8 that's 480 so we have 4 plus four eighty

so this is negative two plus or minus the square root of four hundred and eighty four

the square root of four eighty four is twenty two so now we have negative 2

plus or minus 22 over 16. so now what we're going to do at this

point is separate that into two fractions but let's just uh

let's make some space first so this is negative 2 plus 22 over 16

or negative 2 minus 22 over 16. negative 2 plus 22 that's positive

twenty and twenty over sixteen both numbers are divisible by four

twenty divided by four is five sixteen divided by four is four so the first answer is

five divided by four negative two minus twenty two is negative twenty four

twenty four and sixteen are both divisible by eight negative twenty four divided by eight is

negative 3 16 divided by 8 is 2. and so that's the other answer

negative 3 over 2. so now you know how to use the quadratic formula to solve quadratic equations

you