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Using the Quadratic Formula

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Using the Quadratic Formula

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In this video, I'm going to expose you to what is maybe
one of at least the top five most useful formulas in
mathematics.
And if you've seen many of my videos, you know that I'm not
a big fan of memorizing things.
But I will recommend you memorize it with the caveat
that you also remember how to prove it, because I don't want
you to just remember things and not know
where they came from.
But with that said, let me show you what I'm talking
about: it's the quadratic formula.
And as you might guess, it is to solve for the roots, or the
zeroes of quadratic equations.
So let's speak in very general terms and I'll
show you some examples.
So let's say I have an equation of the form ax
squared plus bx plus c is equal to 0.
You should recognize this.
This is a quadratic equation where a, b and c are-- Well, a
is the coefficient on the x squared term or the second
degree term, b is the coefficient on the x term and
then c, is, you could imagine, the coefficient on the x to
the zero term, or it's the constant term.
Now, given that you have a general quadratic equation
like this, the quadratic formula tells us that the
solutions to this equation are x is equal to negative b plus
or minus the square root of b squared minus 4ac, all
of that over 2a.
And I know it seems crazy and convoluted and hard for you to
memorize right now, but as you get a lot more practice you'll
see that it actually is a pretty reasonable formula to
stick in your brain someplace.
And you might say, gee, this is a wacky formula, where did
it come from?
And in the next video I'm going to show you
where it came from.
But I want you to get used to using it first. But it really
just came from completing the square on this
equation right there.
If you complete the square here, you're actually going to
get this solution and that is the quadratic
formula, right there.
So let's apply it to some problems. Let's start off with
something that we could have factored just to verify that
it's giving us the same answer.
So let's say we have x squared plus 4x minus
21 is equal to 0.
So in this situation-- let me do that in a different color
--a is equal to 1, right?
The coefficient on the x squared term is 1.
b is equal to 4, the coefficient on the x-term.
And then c is equal to negative 21,
the constant term.
And let's just plug it in the formula, so what do we get?
We get x, this tells us that x is going to be equal to
negative b.
Negative b is negative 4-- I put the negative sign in front
of that --negative b plus or minus the
square root of b squared.
b squared is 16, right?
4 squared is 16, minus 4 times a, which is 1, times c, which
is negative 21.
So we can put a 21 out there and that negative sign will
cancel out just like that with that-- Since this is the first
time we're doing it, let me not skip too many steps.
So negative 21, just so you can see how it fit in, and
then all of that over 2a.
a is 1, so all of that over 2.
So what does this simplify, or hopefully it simplifies?
So we get x is equal to negative 4 plus or minus the
square root of-- Let's see we have a negative times a
negative, that's going to give us a positive.
And we had 16 plus, let's see this is 6, 4 times 1 is 4
times 21 is 84.
16 plus 84 is 100.
That's nice.
That's a nice perfect square.
All of that over 2, and so this is going to be equal to
negative 4 plus or minus 10 over 2.
We could just divide both of these terms by 2 right now.
So this is equal to negative 4 divided by 2 is negative 2
plus or minus 10 divided by 2 is 5.
So that tells us that x could be equal to negative 2 plus 5,
which is 3, or x could be equal to negative 2 minus 5,
which is negative 7.
So the quadratic formula seems to have given us
an answer for this.
You can verify just by substituting back in that
these do work, or you could even just try to factor this
right here.
You say what two numbers when you take their product, you
get negative 21 and when you take their sum you get
positive 4?
So you'd get x plus 7 times x minus 3 is
equal to negative 21.
Notice 7 times negative 3 is negative 21, 7 minus 3 is
positive 4.
You would get x plus-- sorry it's not negative --21 is
equal to 0.
There should be a 0 there.
So you get x plus 7 is equal to 0, or x minus
3 is equal to 0.
X could be equal to negative 7 or x could be equal to 3.
So it definitely gives us the same answer as factoring, so
you might say, hey why bother with this crazy mess?
And the reason we want to bother with this crazy mess is
it'll also work for problems that are hard to factor.
And let's do a couple of those, let's do some
hard-to-factor problems right now.
So let's scroll down to get some fresh real estate.
Let's rewrite the formula again, just in case we haven't
had it memorized yet. x is going to be equal to negative
b plus or minus the square root of b squared minus 4ac,
all of that over 2a.
I'll supply this to another problem.
Let's say we have the equation 3x squared plus 6x is equal to
negative 10.
Well, the first thing we want to do is get it in the form
where all of our terms or on the left-hand side, so let's
add 10 to both sides of this equation.
We get 3x squared plus the 6x plus 10 is equal to 0.
And now we can use a quadratic formula.
So let's apply it here.
So a is equal to 3.
That is a, this is b and this right here is c.
So the quadratic formula tells us the
solutions to this equation.
The roots of this quadratic function, I guess
we could call it.
x is going to be equal to negative b.
b is 6, so negative 6 plus or minus the
square root of b squared.
b is 6, so we get 6 squared minus 4 times a, which is 3
times c, which is 10.
Let's stretch out the radical little bit, all of that over 2
times a, 2 times 3.
So we get x is equal to negative 6 plus or minus the
square root of 36 minus-- this is interesting --minus 4 times
3 times 10.
So this is minus-- 4 times 3 times 10.
So this is minus 120.
All of that over 6.
So this is interesting, you might already realize why it's
interesting.
What is this going to simplify to?
36 minus 120 is what?
That's 84.
We make this into a 10, this will become an
11, this is a 4.
It is 84, so this is going to be equal to negative 6 plus or
minus the square root of-- But not positive 84, that's if
it's 120 minus 36.
We have 36 minus 120.
It's going to be negative 84 all of that 6.
So you might say, gee, this is crazy.
What a this silly quadratic formula you're
introducing me to, Sal?
It's worthless.
It just gives me a square root of a negative number.
It's not giving me an answer.
And the reason why it's not giving you an answer, at least
an answer that you might want, is because this will have no
real solutions.
In the future, we're going to introduce something called an
imaginary number, which is a square root of a negative
number, and then we can actually express this in terms
of those numbers.
So this actually does have solutions, but they involve
imaginary numbers.
So this actually has no real solutions, we're taking the
square root of a negative number.
So the b squared with the b squared minus 4ac, if this
term right here is negative, then you're not going to have
any real solutions.
And let's verify that for ourselves.
Let's get our graphic calculator out and let's graph
this equation right here.
So, let's get the graphs that y is equal to-- that's what I
had there before --3x squared plus 6x plus 10.
So that's the equation and we're going to see where it
intersects the x-axis.
Where does it equal 0?
So let me graph it.
Notice, this thing just comes down and then goes back up.
Its vertex is sitting here above the x-axis and it's
upward-opening.
It never intersects the x-axis.
So at no point will this expression, will this
function, equal 0.
At no point will y equal 0 on this graph.
So once again, the quadratic formula seems to be working.
Let's do one more example, you can ever see
enough examples here.
And I want to do ones that are, you know, maybe not so
obvious to factor.
So let's say we get negative 3x squared plus 12x plus 1 is
equal to 0.
Now let's try to do it just having the quadratic formula
in our brain.
So the x's that satisfy this equation are going to be
negative b.
This is b So negative b is negative 12 plus or minus the
square root of b squared, of 144, that's b squared minus 4
times a, which is negative 3 times c, which is 1, all of
that over 2 times a, over 2 times negative 3.
So all of that over negative 6, this is going to be equal
to negative 12 plus or minus the square root
of-- What is this?
It's a negative times a negative so they cancel out.
So I have 144 plus 12, so that is 156, right?
144 plus 12, all of that over negative 6.
Now, I suspect we can simplify this 156.
We could maybe bring some things out
of the radical sign.
So let's attempt to do that.
So let's do a prime factorization of 156.
Sometimes, this is the hardest part, simplifying the radical.
So 156 is the same thing as 2 times 78.
78 is the same thing as 2 times what?
That's 2 times 39.
So the square root of 156 is equal to the square root of 2
times 2 times 39 or we could say that's the square root of
2 times 2 times the square root of 39.
And this, obviously, is just going to be the square root of
4 or this is the square root of 2 times 2 is just 2.
2 square roots of 39, if I did that properly, let's
see, 4 times 39.
Yeah, it looks like it's right.
So this up here will simplify to negative 12 plus or minus 2
times the square root of 39, all of that over negative 6.
Now we can divide the numerator and the denominator
maybe by 2.
So this will be equal to negative 6 plus or minus the
square root of 39 over negative 3.
Or we could separate these two terms out.
We could say this is equal to negative 6 over negative 3
plus or minus the square root of 39 over negative 3.
Now, this is just a 2 right here, right?
These cancel out, 6 divided by 3 is 2, so we get 2.
And now notice, if this is plus and we use this minus
sign, the plus will become negative and the negative will
become positive.
But it still doesn't matter, right?
We could say minus or plus, that's the same thing as plus
or minus the square root of 39 nine over 3.
I think that's about as simple as we can get this answered.
I want to make a very clear point of what I
did that last step.
I did not forget about this negative sign.
I just said it doesn't matter.
It's going to turn the positive into the negative;
it's going to turn the negative into the positive.
Let me rewrite this.
So this right here can be rewritten as 2 plus the square
root of 39 over negative 3 or 2 minus the square root of 39
over negative 3, right?
That's what the plus or minus means, it could be this or
that or both of them, really.
Now in this situation, this negative 3 will turn into 2
minus the square root of 39 over 3, right?
I'm just taking this negative out.
Here the negative and the negative will become a
positive, and you get 2 plus the square root
of 39 over 3, right?
A negative times a negative is a positive.
So once again, you have 2 plus or minus the
square of 39 over 3.
2 plus or minus the square root of 39 over 3 are
solutions to this equation right there.
Let verify.
I'm just curious what the graph looks like.
So let's just look at it.
Let me clear this.
Where is the clear button?
So we have negative 3 three squared plus 12x plus 1 and
let's graph it.
Let's see where it intersects the x-axis.
It goes up there and then back down again.
So 2 plus or minus the square, you see-- The square root of
39 is going to be a little bit more than 6, right?
Because 36 is 6 squared.
So it's going be a little bit more than 6, so this is going
to be a little bit more than 2.
A little bit more than 6 divided by 2 is a little bit
more than 2.
So you're going to get one value that's a little bit more
than 4 and then another value that should be a little bit
less than 1.
And that looks like the case, you have 1, 2, 3, 4.
You have a value that's pretty close to 4, and then you have
another value that is a little bit-- It looks close to 0 but
maybe a little bit less than that.
So anyway, hopefully you found this application of the
quadratic formula helpful.

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Using the Quadratic Formula

Using the Quadratic Formula