This resource is the second
in a two-part series on methods
for solving quadratic equations.
Part One defines quadratic equations.
We then look at a general solving algorithm
supposing that the quadratic can be factored.
We look at multiple examples.
Part Two considers the generic solution method
for quadratic equations, the quadratic formula.
The formula is presented,
then we consider multiple examples
including an application problem.
What happens when we want to solve
a quadratic equation that cannot be factored?
We must use the all-powerful quadratic formula!
To use the formula,
the equation must be written in standard form,
a x squared plus b x plus c equal to zero.
The solutions for x are then equal to
negative b, plus or minus
the square root of the quantity
b squared minus four times a times c,
all over two times a.
This formula looks very imposing.
To use it,
first we find our values for a, b, and c.
We need to be very careful about minus signs.
Then we plug those values into the formula.
We do all the multiplications.
Combine the numbers under the radical.
Take the square root.
Separate the solutions into the plus
and minus cases.
Simplify your answers.
Let’s take a look
at the quadratic formula in action!
Suppose we want to solve the equation
four x squared plus twenty-three x
plus fifteen equals zero.
We may be able to factor this,
but all that guessing and checking will take too long.
Let’s just use the quadratic formula!
First we need to identify our values
for a, b, and c.
A is four,
b is twenty-three
and c is fifteen.
We plug those values into the formula.
We have negative twenty-three
plus or minus the square root of the quantity
twenty-three squared
minus four times four times fifteen,
all over two times four.
We do all the multiplication-type operations.
We are left with negative twenty-three
plus or minus the square root of
five hundred twenty-nine minus two hundred forty,
all over eight.
Combining under the radical,
we have negative twenty-three
plus or minus the square root
of two hundred eighty-nine
divided by eight.
Taking the square root leaves
negative twenty-three plus or minus seventeen
over eight.
Split the plus and minus possibilities.
Negative twenty-three plus seventeen
equals negative six,
divided by eight
simplifies to negative three fourths.
Negative twenty-three minus seventeen
is negative forty,
divided by eight is negative five.
Let’s look at another example
to ensure we know how to use the formula.
What if we have the equation
ten x squared
minus x minus twenty-four equals zero?
In this example, our a is ten.
Our b, the coefficient of the x-term, is negative one.
C is the constant, which is negative twenty-four.
Plugging into the formula, we have x equals
the opposite of negative one,
plus or minus the square root of the quantity
negative one squared
minus four times ten times negative twenty-four,
all divided by two times ten.
Multiplying out,
we have x equals one
plus or minus the square root of one
plus nine hundred sixty,
divided by twenty.
Combining under the radical
we get nine hundred sixty-one.
The square root equals thirty-one,
so we have x equals one
plus or minus thirty-one
over twenty.
Splitting into our plus and minus options,
one plus thirty-one is thirty-two,
which divided by twenty
simplifies to eight fifths.
One minus thirty-one equals negative thirty,
divided by twenty is negative three halves.
But why all this focus
on solving quadratic equations, anyway?
Many real-world phenomena
can be modeled using quadratic equations.
Any projectile motion problem
is in fact a quadratic equation.
So any problem that involves gravity
is really a quadratic equation,
just waiting to happen.
For example,
when a shot put is released
at an angle of sixty-five degrees,
the height of the shot can be modeled
by the equation
negative zero point zero four x squared
plus two point one x
plus six point one,
where x is the horizontal distance
from the point of release.
We want to find the total distance of the throw.
Because we want to find the distance of the throw,
we need to find out how far the shot travels
when it hits the ground.
When the ball is on the ground,
its height above the ground is zero.
So this problem asks us
to set the equation equal to zero
and then solve for x.
Inspecting our equation,
we find that our values for a, b, and c
are negative zero point zero four,
two point one, and six point one, respectively.
We plug those values into the quadratic formula.
Do all the multiplication-type operations.
Combine the quantities under the radical.
Take the square root.
Now we split the equation
into its plus and minus options and simplify.
Looking at the plus option,
we have the quantity negative two point one
plus two point three two one
divided by negative zero point zero eight,
which is negative two point seven six two five.
For the minus option,
we have the quantity negative two point one
minus two point three two one
divided by negative zero point zero eight,
which is fifty-five point two six two five.
Only the positive answer makes any sense.
The negative option is what happens
if the shot putter drops the ball
and it goes backwards.
The correct distance of the throw
is a little bit more
than fifty-five feet three inches.
The quadratic formula is the default
generic solution method for any quadratic equation.
One reason it is so important
to be able to solve quadratics
is that they model any number
of real-world phenomena,
especially any projectile motion.
Any problem that involves gravity
is likely to be some form of quadratic equation.
To use the formula,
you must ensure that all the values
are on one side of the equation
with zero on the other side.
You then identify the a, b, and c
as the coefficients of the squared,
linear, and constant terms, respectively.
Those values are entered into the formula.
Work through all the multiplications,
then combine the values under the radical.
Take the square root
then simplify the solutions
into the plus and minus options.
Simplify, and you have found your answer!
