- Using the quadratic formula
to solve quadratic equations.
The solution of a quadratic
equation AX squared + BX + C = 0
are given by the quadratic
formula, X = -B plus or minus
the square root of B squared -
4AC divided by 2A.
The quadratic formula
can be used to solve
any quadratic equation.
When using the quadratic
formula, for the fraction part
in the quadratic formula,
notice that the term -B is part
of the numerator.
You want to be very careful
about how you write this.
The correct form is -B
plus or minus the square root
of B squared - 4AC
all over 2A.
The incorrect form is to
write -B plus or minus
then your fraction.
That is to be avoided.
In this example use the
quadratic formula to solve
the equation, all solutions
are real numbers.
5x squared + 5X - 1 = 0.
To use the quadratic formula
first you need to remember
that AX squared + BX + C = 0,
that means A is 5, B is 5,
C is -1.
Then using our quadratic
formula, X = opposite of 5,
which is -5, plus or minus
the square root of 5 squared
- 4 x 5 x -1, our square root
symbol should cover all of that,
all of these is
divided by 2 x 5.
Once you have substituted
into the formula,
you can do the math to find X.
So I have -5 plus or minus,
5 squared is 25 = 4 x 5 x 1,
is 20 divided by 10,
which is -5 plus or minus the
square root of 45 divided by 10.
-5 plus or minus,
45 is not a perfect square,
but part of 45 is
a perfect square.
Since 45 is 9 x 5 for
the square root of 45,
I could take the square
root of 9 which is 3,
and leave my square root of 5,
and that's divided by 10.
If we were to reduce this,
it's very important to remember
that the 5, the 3, and the 10
all have to have something
in common, so while 5 and 10
have a common faction 5,
5, 10, and 3 do not have
anything in common
so this answer cannot
be reduced.
So X = -5 plus or minus 3
square root of 5 divided by 10,
or we can write the answer as
X = -5 + 3 square root of 5
divided by 10, or -5 - 3
square root of 5/10.
In this example, use
the quadratic formula
to solve the equation, all
solutions are real numbers,
5X squared - 4X = 2.
So we need our quadratic
equation to be in the format
of AX squared plus BX + C = 0.
So the first step here is to set
this equal to zero,
5X squared - 4X - 2 = 0.
So, now I can pick out
my A which is 5,
my B which is -4,
and my C which is -2.
Now, I'm ready to use my
quadratic formula to find X.
So I bring down my X =
the opposite of -4,
so that's a positive 4,
plus or minus the square root
of -4 squared - 4 x A,
which is 5, x C,
which is 2, all divided
by 2 x A, which is 5.
So now we can do our math.
So I have 4 plus or minus
the square root of
16 + 40 divided by 10,
which equals 4 plus or minus the
square root of 56 divided by 10.
So 56 can be written as 4 x 14,
4 plus or minus the square root
of 4 is 2, square of 14,
we don't know, divided by 10.
So when I try to decide if I can
reduce, I need to look at the 4,
the 10, and the 2, not the 14.
Since all three are even,
I can divide 4 by 2,
which gives me 2 plus or minus
2 by 2 which gives me 1,
square root of 14 divided
by 10 divided by 2 which is 5.
So my answer is X = 2 plus or
minus the square root of 14
all over 5, or I can write my
answer as 2 + the square root
of 14 divided by 5, or 2 -
the square of 14 all over 5.
And remember that this
is our X value,
so our X equals is important.
In this example we're going
to use the quadratic formula
to solve our equation,
5/AX squared + 3/4X = 1/4.
It's going to be cumbersome
to use fractions
in our quadratic formula.
If I were to multiply this
equation by the least common
multiple of 8, 4, and 4,
I can get rid of my fractions.
I can only do something like
this when I have an equality.
The smallest number that will
cancel an 8 and a 4 would be 8.
I would have (8/1) x (5/8)
x squared + (8/1) x (3/4)X
= (8/1) x (1/4).
My 8s would cancel leaving
me with 5X squared,
plus my 8 and my 4 cancels,
2 x 3 is 6, and then my X =,
my 8, and my 4 cancel, leaving
me with 2 x 1, which is 2.
And then getting it in the form
of AX squared + BX + C = 0,
I would subtract the 2, so
I have 5X squared + 6X - 2 = 0,
and now I'm ready to use
my quadratic formula.
So now we can find A
which is 5, B which is -6,
C which is -2.
Substituting into our quadratic
formula, X = the opposite of -6,
so positive 6 plus or minus
the square root of B square,
so it's -6 squared - 4 x A,
which is 5, x C, which is -2,
all divided by 2 x A,
which is 5.
So that equals 6 plus or minus
the square root,
-6 squared is 36, -4 x 5 x -2 is
a positive 40 all over 10.
That equals 6 plus or minus
the square root of 76/10,
76 is even, so calculating.
So I have 76, it's even,
so I'm going to divide it by 2,
and I get 38.
Still even so I'm going
to divide it by 2 again,
and I get 19.
So I can write 76 as 19 x 4.
I'll be able to take
the square root of 4 but not
the square root of 19.
So we have 6 plus or minus,
the square of 4 is 2,
square root of 19, we don't
know, all over 10.
So that is going to reduce
because 6, 2, 10 are all even.
6 divided by 2 is 3, plus or
minus, 2 divided by 2 is 1,
square root of 19,
and 10 divided by 2 is 5.
So my final answer is
X = -3 plus or minus
the square root of 19
all over 5.
I can also write my X value as
-3 + the square root of 19/5,
or -3 - the square root
of 19 all over 5.
In this example find
all X intercepts,
F(X) = -5 X squared + 3X + 2.
Because we're looking for
the X intercept that means
F(X) is 0, so 0 = -5X
squared + 3X + 2.
And our quadratic equation is
in the format for us to find
A, B, and C, and then
use our quadratic formula.
So A is -5, B is 3, and C is 2.
The nice thing about
the quadratic formula is,
it doesn't matter if you're
leading term is positive
or negative.
If I where factoring,
I would need my leading term
to be positive, so I'd have to
factor out a negative one,
with the quadratic formula,
that doesn't matter.
So I'm ready to substitute
x = the opposite of 3,
so -3, plus or minus the square
root, so 3 squared - 4 x -5 x 2
all divided by 2 x -5.
So X = -3 plus or minus
the square root,
3 squared is 9, - 4 x -5 x 2
is a positive 40
all over 2 x -5 which is -10.
So this equals -3 plus or minus
the square root of 49
all over -10.
The square root of 49 is
a perfect square so I have
-3 plus or minus 7/-10.
Unlike the other problems we
have done with the quadratic
formula, this one can actually
be simplified further.
So we need to write this as
X = -3 + 7/-10 and -3 - 7
all over -10.
Then we simplify, so -3 + 7
is 4/-10, 4/10 reduces to -2/5.
Now looking at -3 - 7/-10,
-3 - 7 is -10,
divided by -10 reduces to 1.
So my X intercepts are (-2/5,0),
and (1,0).
Determining the number of real
number solutions.
When you're deciding how you
want to solve your quadratic
equation, you want to try to
decide what's going to be
the easiest to use.
Do I want to factor?
Do I want to
complete the square?
Do I want to use
the quadratic formula?
To help you decide that,
it's useful to know what type
of solution you're
going to have.
For the quadratic equation,
AX squared + BX + C = 0,
the discriminant is
B squared - 4AC.
Also if B squared - 4AC
is greater than 0
that means there are two
real number solutions.
The two real number solutions
may or might not include
a square root.
If B squared - 4AC = 0, there
is one real number solution.
This is going to be when you
have a perfect square trinomial.
If B squared - 4AC
is less than 0,
then there are two imaginary
solutions and no real number
solutions.
In this example determine the
number and type of solutions
of the equation, which means
we don't actually have
to solve the equation, we just-
are trying to decide what
our solutions will look like.
5X squared - 3X + 2 = 0,
so using the discriminant,
I want B squared so that's
-3 squared - 4 x A,
which is 5, x C, which is 2,
that equals 9 - 4 x 5 is 20,
x 2 is 40, and 9 - 40 is -31.
Since my 31 is negative, I have
two imaginary number solutions.
Example, determine the number
and type of solutions
of the equation,
X squared + 6X + 9 = 0.
So using my discriminant,
I want to substitute B squared,
B is 6, so 6 squared - 4 x A,
which is 1, x C, which is 9.
This gives me 36 - 36
which equals 0,
so I have one real
number solution.
In this example use the
quadratic formula to find
all complex number solutions.
X squared - 3X + 15 = 0.
I have X = the opposite of -3,
so that will be 3,
plus or minus the square root
of -3 squared - 4 x A,
which is 1, x C, which is 15,
that's all divided by-
so 2 x 1.
So this gives me 3 plus or minus
the square root of 9 - 60
divided by 2 which is 3 plus or
minus the square root of -51/2.
So X = 3 plus or minus i square
root of 51 all over 2,
which I can write as
3 + i square root of 51
divided by 2, or 3 - i square
root of 51 all over 2.
In this example use the
quadratic formula to find
all complex number solutions.
2(5X - 2) = -1.
So we're going to have to do
some math here to get this
in our format of
AX squared + BX + C = 0.
So first I'm going
to distribute my 2X
so I have 10X squared - 4X,
then I'll add one to both sides,
+ 1 = 0.
I'm ready for my quadratic
formula now,
and I can substitute.
X = the opposite of -4,
which is 4,
plus or minus the square root
of -4 squared - 4 x 10 x 1
all over 2 x 10.
This equals 4 plus or minus
the square root of 16 - 40
all over 20.
4 plus or minus the square root
16 - 40 is -24 all over 20.
I can simplify 24
because it is 4 x 6,
and I can simplify my negative
to i, so I have 4 plus or minus,
the square root of 4 is 2,
the square root of -1 is i,
6 stays under the root
divided by 20.
Because 4, 2, and 20 are even
I can reduce this further.
4 divided by 2 is 2, plus
or minus 2 divided by 2
is 1i square root of 6 divided
by 20, divided by 2 is 10.
So my final answer is X = 2 plus
or minus i square root of 6
all over 10,
or I can write my answer X =
2 + i square root of 6/10,
or 2 - i square root of 6/10.
Using the quadratic formula to
solve quadratic equations,
deciding which method to use to
solve a quadratic equation,
if our method is factoring when
to use would be for fractions
that can easily be put into the
form AX squared + BX + C = 0,
and where AX squared + BX + C
can be easily factored.
The square root property are for
equations that can be easily
put into the form X squared = K,
or (X + P) squared = K.
Completing the square, use when
the directions require it.
Quadratic formula, for all
equations except those
which can be easily solved
by factoring or by using
the square root property.
