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in this lesson we're going to talk about
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solving quadratic equations by factoring
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so let's start with this example x
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squared minus 49 is equal to zero
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you can use the difference of perfect
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squares technique for this one
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the square root of x squared is x
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the square root of 49 is seven
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so it's going to be x plus seven
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and x minus seven
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now you need to set each factor equal to
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zero at this point and then you could
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find the value of x
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so we have x plus seven is equal to zero
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and x minus seven is equal to zero
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the reason why we can do that is because
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if one of these terms is equal to zero
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then everything is zero zero times
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anything is zero
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so x
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is equal to negative seven
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and in the other equation if we add
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seven to both sides we could see that x
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is equal to positive 7.
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let's try another example
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let's say if we have 3x squared
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minus 75
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is equal to zero
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what is the value of x
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3 and 75
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are not perfect squares
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so we don't want to use the difference
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of perfect squares technique yet however
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we can take out the gcf the greatest
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common factor
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which is three three x squared divided
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by three
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is x squared
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negative seventy-five divided by three
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is negative twenty-five
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now we can use the difference of perfect
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squares technique
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to factor x squared minus 25.
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the square root of x squared is x the
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square root of 25 is 5.
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so it's going to be x plus 5 and x minus
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5.
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so if we set x plus five equal to zero
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we can clearly see that x
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will be equal to negative five and if we
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set x minus five equal to zero
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x is equal to plus five
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and so that's it for that one
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now what about this one let's say if we
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have 9x squared
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minus 64
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is equal to zero
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well first
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we can use the difference of perfect
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squares technique we can square root 9
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and we can square root 64.
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the square root of 9 is 3.
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the square root of x squared is x
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the square root of 64 is 8.
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so it's going to be 3x plus 8 3x minus
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8.
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so if we set 3x plus 8 equal to 0
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then we can see that 3x is equal to
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negative 8 which means x is equal to
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negative eight over three
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now if we set three x minus eight equal
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to zero and solve for x
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x is gonna be positive eight over three
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using the same steps
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now what if we have a trinomial
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x squared minus 2x minus 15.
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and the leading coefficient is one how
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can we factor this expression
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all you need to do is find two numbers
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that multiply to negative 15 but that
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adds to negative two
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numbers that multiply to fifteen are
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five and three
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so we have positive five and negative
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three or negative five and three
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five plus negative three adds up to
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positive two but negative 5 plus 3 adds
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up to negative 2.
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so this is what we want to use
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it turns out that
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to factor it it's simply going to be x
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minus
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5 plus
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x plus 3.
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so if we set x minus five equal to zero
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x will be equal to five
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and if we set x plus three equal to zero
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x will be equal to negative three
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let's try another one like that
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let's say if we have x squared plus
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3x
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minus 28
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so what two numbers multiply to negative
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28 but add to three
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go ahead and try it
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so if we divide 28 by 1
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we'll get negative 28 if we divide
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negative 28 by 2
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negative 14 3 doesn't go into it if we
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divide it by 4 we'll get negative 7. 4
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and negative 7 differs by three if we
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add them it's negative three
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so we need to change the sign
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so it's going to be x minus four times x
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plus seven
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which means that x
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is equal to positive four
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and negative seven
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here's another problem
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so how can we factor this trinomial when
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the leading coefficient is not one
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so what we need to do in this problem we
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need to multiply eight and negative
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fifteen
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eight times negative 15
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is negative 120.
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now what two numbers multiply to
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negative 120 but add to two
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if you're not sure make a list
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let's start with one we have one in 120
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two and sixty
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three and forty
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four and thirty
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five and twenty four
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six and twenty
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eight and fifteen
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now 10 and 12 seem promising
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10 and negative 12 differ by negative 2
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but positive 12 and negative 10
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adds up to positive 2.
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so what we're going to do in this
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problem is we're going to replace 2x
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with 12x and negative 10x
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and then factor by grouping
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in the first two terms let's take out
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the gcf
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which is going to be 4x
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8x squared divided by 4x
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is 2x
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and 12x divided by 4x
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is 3.
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and the last two terms
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take out the greatest common factor in
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this case negative 5.
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negative 10x divided by negative 5
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is 2x
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negative 15 divided by negative 5 that's
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plus 3.
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now if you get two common terms
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that means you're on the right track you
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can write it once
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in
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a parenthesis in the next line
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now the stuff on the outside 4x and
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negative 5 that's going to go in the
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second parentheses
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so that's what we have
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now let's set two x plus three equal to
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zero
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and 4x minus 5 equal to 0.
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so in the first equation let's subtract
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3 from both sides
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so 2x is equal to negative 3.
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and then let's divide by 2.
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so the first answer x is equal to
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negative three over two
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now let's find the other answer
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so let's add five to both sides
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so we can see that four x is equal to
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five
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and then let's divide both sides by four
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so x
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is equal to five over four
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and that's it for this problem
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now let's get some of the answers
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to the quadratic equations that we had
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in the last lesson
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so for this particular problem
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when we factor it we got a solution of 5
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and negative 3
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in less than 10.2
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but now let's use the quadratic equation
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to get those same answers
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so x is equal to negative b
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plus or minus the square root of b
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squared minus 4ac
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divided by 2a that's the quadratic
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formula
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and you need the quadratic equation in
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standard form
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so we can see that a is equal to one
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b
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is the number in front of x b is
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negative two
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and c
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is negative fifteen
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so let's replace b with negative two
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b squared or negative two squared
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negative two times negative two is four
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a is one and c is negative fifteen
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divided by two a or two times one which
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is two
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negative times negative two is positive
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two
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and then we have four negative four
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times negative fifteen that's positive
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sixty
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and sixty plus four is sixty-four
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now the square root of sixty-four is
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eight
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so we have two plus or minus eight
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divided by two
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two plus eight is ten
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ten divided by two is five that gives us
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the first answer the next one is two
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minus eight divided by two
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two minus eight is negative six negative
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six divided by two is negative three
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which gives us the second answer
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so you can solve a quadratic equation by
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factoring
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or by using the quadratic formula
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now let's try another example
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eight x squared plus two x
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minus fifteen
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use the quadratic equation to find the
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values of x
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so we can see that a is equal to eight
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b is the number in front of x that's two
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c is negative 15.
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so using the quadratic formula x equals
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negative b
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plus or minus the square root
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of b squared minus 4ac
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divided by 2a
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so b is 2
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which means b squared that's going to be
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positive 4
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minus 4 times a a is 8 c is negative 15
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divided by 2 a or 2 times 8 which is 16.
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so this is negative 2 plus or minus
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square root
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4.
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now negative 4
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times
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negative 15 is positive 60 60 times 8
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that's 480
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so we have 4 plus four eighty
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so this is negative two plus or minus
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the square root of four hundred and
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eighty four
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the square root of four eighty four is
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twenty two
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so now we have negative 2
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plus or minus
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22 over 16.
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so now what we're going to do at this
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point is separate that into two
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fractions
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but let's just uh
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let's make some space first
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so this is negative 2
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plus 22 over 16
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or
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negative 2 minus 22 over 16.
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negative 2 plus 22 that's positive
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twenty
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and twenty over sixteen
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both numbers are divisible by four
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twenty divided by four is five
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sixteen divided by four is four
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so the first answer is
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five divided by four
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negative two minus twenty two
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is negative twenty four
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twenty four and sixteen are both
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divisible by eight
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negative twenty four divided by eight is
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negative 3
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16 divided by 8 is 2.
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and so that's the other answer
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negative 3 over 2.
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so now you know how to use the quadratic
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formula to solve quadratic equations
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you