This video is for Section 8.2
our objectives in this section are to solve quadratic equations using the quadratic formula
and use the discriminant to determine the number and type of solutions
so we're going to start with the quadratic formula
quadratic formula states that the solutions of a quadratic equation
when it's in the form a x squared plus bx plus c equals zero
so just like with factoring we need to have zero on one side of the equation
are given by the quadratic formula
which is x equals negative b plus minus square root of b squared minus 4ac, all over 2a
so what we're going to do is we're gonna plug in the numbers for a, b and c, the coefficients in the equation
and that's gonna give you our two values for x
now the plugging in is the easy part with these
the simplifying is the part that's a little bit tricky, especially with MyMathLab
because they're gonna want things to be completely simplified
and so, even if you have the right answer
things are gonna be marked wrong if they're not completely simplified
so example one, we're gonna look at first, it says to solve using the quadratic formula
in this equation, we already have zero on one side
so that is not something we need to do, we can just move on to plugging directly into the formula
so I'm gonna label a, b, and c
and then I'm going to plug in here, so it's gonna be negative b
b is two, plus minus
square root of b squared minus four times a times c
if there's a negative sign in front of the variable, then that needs to be included as well
and then everything divided by 2a
so once I have everything plugged in, what I'm gonna do is simplify what's under the square root first
and get that down to a single number, then that's going to tell me what other simplifying steps I may need to do
so 2 squared is 4
and, now you have to be careful because there's a minus sign and then a minus sign here
so minus 4 times 8 times negative 1, those two minus signs combine to make a plus
even though they're kind of far away from each other, those two minus signs
minus, are gonna be minus negative and then that's going to make a positive
so that's one of the first, common ways to make a mistake is to get the sign wrong at that step
I'm gonna do 2 times 8 here and get 16
and now I'm going to add those two numbers under the square root together, 4 plus 32 is 36
so what I have here, is I have the square root of 36, which is a perfect square
so, sometimes when you're at this step, you can't simplify anymore
sometimes you can, and in this case I can because 36 is a perfect square
so if you try to enter this into MyMathLab, even though all the work that we've done so far is correct
MyMathLab is gonna mark it wrong because it's not fully simplified
so square root of 36 is 6
and from here, what I'm gonna do is split this up, so I get two answers
so I'm going to do one with the plus sign, negative 2 plus 6 over 16
and one with the minus sign, negative 2 minus 6 over 16
so, if the square root simplifies to a whole number
then you're gonna split it up and get 2 answers, if it doesn't simplify to a whole number
then you can just leave plus minus with the square root
but if it is a perfect square, then you're gonna need to actually take the square root
and then get two separate answers, so negative 2 plus 6 is 4
this is gonna be 4 over 16, which is 1/4
negative 2 minus 6 is negative 8, negative 8 over 16 is negative 1/2
you are also gonna need to, want to reduce fractions, because if you put in a fraction that can be reduced
MyMathLab is gonna mark it wrong
so my two answers here are 1/4 and negative 1/2 and in MyMathLab
just like we saw earlier with these problems
they're going to have a bracket, you're gonna type the answers in the bracket separated by a comma
do not put parentheses around them, it's not an ordered pair, these are two possible solutions for x
so it's more like a list of solutions, so you type it in just like that
with the fractions, they're going to want you to enter the fractions and not decimal approximations
so just be aware of that as well
so that's my answer for example one
so next you should try checkpoint problem number one
do this in your notes, check your answer, and then move on with the rest of the video
next we're gonna look at example two, it says solve using the quadratic formula
in this case, we do not have zero on one side of the equation
so that's going to be my first step, is to put zero on one side of the equation
I'm going to move the 4x and the 1 over
although with this, keeping a leading coefficient positive is not quite as important
but I usually like to do that anyway, just out of habit
so that would give me 2x squared minus 4x minus 1
if you moved the 2x squared over, and had minus 2x squared plus 4x plus 1
it should come out to be the same answer in the end, so you can do it that way also
so this is going to be my equation, 2x squared minus 4x minus 1 equals 0, and I'm going to label a, b and c
and then plug in, now when I'm plugging in this one
it's minus b, if b is negative
that's gonna be minus negative 4, which is going to change to plus 4
so another place where you can make a little sign mistake that gives you the wrong answer
so negative b, negative negative 4 is plus 4
and it's gonna be b squared, when you're squaring this even though it's negative
it should come out to be positive, negative times negative is positive
minus 4 times a times c, all over 2a
so start by working with what's under the square root
positive 4, plus or minus
negative four squared is positive sixteen, and then this is gonna be plus 8
4 times 2 times negative 1, and since it's subtracted, again, minus negative changes to positive
and then we're gonna have 4, plus minus, 16 plus 8 is 24
so what happens with this one, I don't get a perfect square under the square root, 24 is not a perfect square
however, it is a root that can be simplified
and again, if it can be simplified you need to do that, if it couldn't be simplified then I would just be done here
so if it was square root of 23, for example
there's nothing that that can be divided by that's a perfect square, so the problem would just be done
but since it's 24, 24 is divisible by 4
that means it can be simplified, since 4 is a perfect square
so square root of 4 times square root of 6
square root of 4 is 2
so I can simplify that to 2 square root of 6
now, there's one last thing that I can do to simplify here
and that's looking at these 3 numbers
before here, the number in front of the root, not the number underneath the root, and the denominator
if there's a number that all three of those numbers can be divided by, that can be taken out
so in this case, there is such a number
they can all be divided by 2
so the fours, I'm going to divide by 2 and get 2, the 2 I'm going to divide by 2 and get 1
so this is going to be 2, plus or minus
1 times square root of 6, I can just write a square root of 6
and then put a 2 on the bottom
so this is the answer
again, you're looking at the two numbers on the top, not the number under the root, and the denominator
if there's something that they all can be divided by, that can be taken out
so next you should try checkpoint problem number two
do this in your notes, check your answer, and then move on with the rest of the video
lastly we're going to look at example three
again, I don't have 0 on one side so I'm gonna start by putting it in that form
so I'm gonna add 4x to both sides
and when I do that, I'm going to make sure that I write it in the middle
so that I identify the correct values of a, b, and c
so a is 3, b is 4, c is 2
so go ahead and plug into the quadratic formula
negative b, plus minus, square root
b squared is 4, minus 4, whoops, 4
times a times c
all divided by 2 times a
so start by working on what's under the square root, 4 squared is 16
4 times 3 times 2, is 12 times 2, which is 24
now in this case, I don't have any negatives
so that minus sign is gonna stay a minus sign
because I don't have any, 4 times 3 times 2 is positive 24
so 16 minus 24 is negative 8
now, simplifying the square root of negative 8
this is going to use imaginary numbers, so we're using all the stuff that we used back in chapter 7
what I'm gonna do is use the i
that's gonna be i times square root of positive 8, 8 can also be simplified because it's 4 times 2
so I would write this as 2i square root of 2
so we have negative 4, plus minus 2i square root of 2, over 6
now looking at these three numbers, they can all be divided by 2
4, 2, and 6
so this is going to become 2, this is going to become 1, this is going to become 3, divide them all by 2
and that leaves me with my answer 2, plus minus i square root of 2, over 3
the book shows breaking this up into separate fractions, which you can also do
but if you enter it this way into MyMathLab, that'll be fine
it'll accept that for the correct answer
so, next you should do checkpoint problem number 3
do this in your notes, check your answer, and then move on to the homework
