Hi, this is Walter Stone, and
this lesson is Unit 9, Lesson 2.
It's about the quadratic formula.
And after finishing this lesson,
you should be able to solve quadratic
equations by using the quadratic
formula and use the discriminant
to determine the nature of
the roots of a quadratic equation.
And the word roots is another way of
saying solutions to a quadratic equation.
So, recall that a quadratic
equation is an equation of the form
x squared + bx + c = 0,
where a is unequal to 0
cuz we do want this x squared
term to be part of the equation.
So, one question that gets asked
a lot in an algebra course like this,
is where does the quadratic
formula come from?
And one way of looking at getting
a solution to this equation
is using the process of
completing the square.
So in the prior lesson, we discussed that
in order to solve by
completing the square,
we have to have a leading
coefficient of 1.
So, one thing we can do is
we can divide every term by
a in order to get the leading
coefficient to be 1.
And then we can subtract c/a to bring
the constant term to the right hand side.
And then in order to complete the square,
we take half of the coefficient of x and
we square it.
So half of b/a is b/2a, and
we're going to square this number.
And we're going to add
that number to both sides.
So on the left hand side,
completing the square means we can
write the left hand side
as (x + b/2a) squared.
And on the right hand side,
this will be b squared/4a squared.
So what happens is when we
simplify the left hand side,
if you have a common denominator of
4a squared, you'd have to multiply
the first fraction here by 4a/4a.
So we would have -4ac/4a squared.
And then we'll have b squared/4a squared.
So, what we usually do is we
write the positive number first,
so that's why we have b
squared-4ac / 4a squared.
So, what we then do is use
the square root property, and
we take the square root of both sides.
We have x + b/2a on the left hand side,
and we'll have +/-
the square root of b squared- 4ac
/ the square root of 4a squared.
And the square root of 4a squared is 2a.
And then what we can do is we can
subtract b/2a from both sides,
and we'll have -b/2a +/- the square
root of b squared-4ac / 2a.
And then we have a common denominator,
and we can write then -b +/- the square
root of b squared -4ac / 2a, and this
result gives us the quadratic formula.
So if we're solving an equation of
the form, ax squared + bx + c = 0,
the solution will be x =
-b +/- the square root
of b squared-4ac / 2a,
as long as a is not 0.
We want this equation to
be a quadratic equation.
So in the examples we do, we're going
to be using the quadratic formula
to solve these equations.
So suppose we want to solve
x squared + 7x = 3x- 1.
The big deal here is that we have to
have the equation in standard form.
We have to have it in the form
ax squared + bx + c = 0.
And we don't have 0 on
the right hand side.
So what we can do is we can
subtract 3x from both sides.
7x-3x is 4x, and
we can add 1 to both sides.
So in standard form, the original equation
is equivalent to x squared + 4x + 1 = 0.
So here, a = 1, b = 4,
and c = 1, and we can substitute
those values into the formula.
So we'll have the opposite,
what I like to do is,
I like to set up kind of a template
that I can plug these values into.
And I find it helpful,
because sometimes I have to make sure
that I'm evaluating correctly in
terms of the signs of the numbers.
So I always put parenthesis for
b, for b squared, for a, and for
c, and for a in the denominator
to fill in with these values.
So we'll have -(4) +/- the square root of
(4) squared- 4(1)(1) / 2(1),
and we'll have 16-4 is 12.
So we'll have -4 +/-
the square root of 12.
12 is 4 times 3, so
we'll have 2 times the square root of 3,
will be the simplification of root 12.
And both of these numerator
terms are even, so
we can divide both by 2, and
we'll get -2 +/- root 3.
So, try solving the second equation
using the quadratic formula,
and check your work.
So here,
you can see I started you off, and
here, a = 4, b = -7, and c = 3.
And I set up my template to fill in,
and I made a little mistake,
but this -7,
I made sure I substituted that correctly.
a = 4, c = 3, so we have the opposite
of b is the opposite of -7,
+ (-7) squared- 4(4)(3) / 2(4).
And here we have -4(4)(3) -48,
so we have 49- 48 = 1.
And the square root of 1 is 1.
So we'll have 7 +/- 1 / 8.
So this gives us two
calculations we wanna do.
If I take 7 + 1, I get 8/8,
which gives me 1.
And 7-1 is 6, 6/8 is 3/4,
so the solutions will be 1 or 3/4.
So here, since we get these nice
fraction and whole number solutions,
it could be that we could've factored
this on the left, and so it might
be easier to make the choice sometimes to
factor the left hand side if possible.
But using the quadratic
formula will always work.
Try this equation,
solve by using the quadratic formula,
3x squared- 33 = 0, and shut off the video
and come back and check your work.
So here, we have a 0 term here,
the x term is missing.
So we can think of this
as 3x squared + 0x- 33.
And 3 and 33 have a common factor of 3,
so I even divided every term by 3,
just to make things a little bit easier.
And this is like, if we just take
a little side trip here, this is like
the equation x squared- 11 that can be
solved using the square root property.
So we should get +/- root 11 as our
solutions using the quadratic formula.
So if we use the quadratic formula,
a = 1, b = 0, and c = -11.
So if we take the opposite of b,
we have -(0) +/-,
(0) squared-4(1)(11) / 2.
So underneath the radical,
we get positive 44, so
we'll have +/- root 44/2.
And 44 is 4 times 11, so
this will be simplified to
+/- 2 root 11 / 2, the 2s cancel, and
we do get the two solutions of root 11 and
-root 11.
So here, we're just seeing that there
are many ways to solve these equations,
now that we're building kind
of a toolbox to do these.
But the quadratic formula is
something that will always work.
The second objective of
this lesson was to use
the discriminant to determine the nature
of solutions of a quadratic equation.
The discriminant is that value,
b squared-4ac that is
underneath the radical symbol.
And we can use this to describe
the solutions of a quadratic equation.
So we're looking at the nature
of solutions of a quadratic.
So if b squared-4ac is a perfect square
and a positive number, what will happen is
we will get an integer,
we'll have plus or minus an integer.
And so what will happen is we'll get
two different rational solutions for
that equation.
If the discriminant is a positive number
that's not a perfect square, we'll have
a number plus or minus the square
root of a simplified radical.
And so what will happen is, underneath the
radical, we won't have a perfect square.
And those numbers that are square roots
of non-perfect squares are irrational.
So that's why we'll get two
different irrational solutions.
If the discriminant is 0,
then we'll just have the opposite of b/2a,
and that's a rational number,
that's a fraction.
And if the number underneath
the radical symbol is negative,
that's not a real number, so
we'll have no real solutions.
Later, we'll have experiences where
we talk about complex numbers,
and if the discriminant is negative,
we get complex conjugate solutions,
so you may have heard
about that in the past.
But right now, we're just working with the
idea that we won't have real solutions.
So if I want to know the nature
of solutions of this equation,
we need to write the equation
in standard form.
So that's something to look out for
as we've done in some prior examples.
So we can subtract 20x from both sides,
and we'll have 4x squared- 20x + 25 = 0.
So a = 4, b = -20, and c = 25.
And when we take b squared- 4ac,
we'll have (-20) squared- 4(4)(25).
So we'll have 400- 400, which is 0.
So because we get 0, we know that the
equation will have one rational solution.
And another thing that we see here,
another interpretation
of this is this will be a trinomial
square if that happens.
So, it looks like this can
factor into (2x-5) squared.
And so we'll have 5/2 as
a solution to this equation.
And we can check that with
the quadratic formula as well.
Okay, so that's another interpretation,
is if we have 0,
we have a perfect square trinomial.
So we can do a lot of
analysis with these ideas.
So try stating the nature of the solutions
of each of the three equations in
the Try This exercise, pause the video,
and then come back and check your work.
So in the first equation, we had to
make sure it's written in standard form.
It turns out that b squared- 4ac,
that's gonna be (-3)
squared -4(2)(4),
that's 9-32 which is -23.
So since that's a negative number,
this equation will have no real solutions.
In part b,
the equation is in standard form,
and a = 4, b = -5 and c = -2.
b squared -4ac turns out to be 56.
56 is not a perfect square, so we'll
have two different irrational solutions.
And in part c, a = 2, b = -9, c = 4.
So, we'll have (-9) squared- 4(2)(4), and
we'll have 81- 32, which is 49,
and 49 is a perfect square.
So, we'll have two different rational
solutions for that equation.
Okay, so you can try solving each equation
and check that this theory is correct.
So we could find
the discriminate first and
then try solving the equations, okay?
So try the online exercises for this
section, and we will see you in class.
Take care.
