Hello. I'm Professor Von Schmohawk and welcome to Why U.
Hello. I'm Professor Von Schmohawk and welcome to Why U.

In the previous lecture, we showed geometrically
In the previous lecture, we showed geometrically

that any quadratic expression of the form "x-squared, plus bx, plus b over two squared"
that any quadratic expression of the form "x-squared, plus bx, plus b over two squared"

is a perfect square.
is a perfect square.

In this geometric proof
In this geometric proof

we represented the x-squared term
we represented the x-squared term

by the area of a square with sides of length x
by the area of a square with sides of length x

the bx term
the bx term

by the area of a rectangle with sides of lengths b and x
by the area of a rectangle with sides of lengths b and x

or alternatively, two smaller rectangles each with half that area
or alternatively, two smaller rectangles each with half that area

and the constant term "b over two" quantity squared
and the constant term "b over two" quantity squared

by the area of a square with sides of length "b over two".
by the area of a square with sides of length "b over two".

The square created from the sum of these areas
The square created from the sum of these areas

therefore represented the entire quadratic expression.
therefore represented the entire quadratic expression.

And since that square had sides of length "x, plus b over two"
And since that square had sides of length "x, plus b over two"

we saw that this expression can also be written as "x, plus b over two" quantity squared.
we saw that this expression can also be written as "x, plus b over two" quantity squared.

However, this geometric proof assumes that the bx term has a positive value
However, this geometric proof assumes that the bx term has a positive value

and so in the diagram, the area of the two rectangles that together represent bx
and so in the diagram, the area of the two rectangles that together represent bx

are added to the other areas rather than subtracted.
are added to the other areas rather than subtracted.

So what would this proof look like if the value of bx is negative?
So what would this proof look like if the value of bx is negative?

Just as before, we can represent the x-squared term
Just as before, we can represent the x-squared term

as a square with sides of length x
as a square with sides of length x

the "b over two, squared" term
the "b over two, squared" term

as a square with sides of length "b over two"
as a square with sides of length "b over two"

and the bx term
and the bx term

as two rectangles with sides of lengths "b over two" and x.
as two rectangles with sides of lengths "b over two" and x.

However, since the bx term is negative
However, since the bx term is negative

the area of those two rectangles must be subtracted from the other areas instead of added.
the area of those two rectangles must be subtracted from the other areas instead of added.

So let's first subtract the area of the rectangle on the right from the x-squared box.
So let's first subtract the area of the rectangle on the right from the x-squared box.

This leaves a rectangle of height x and width "x, minus b over two".
This leaves a rectangle of height x and width "x, minus b over two".

Next, let's move the small square with a side length of "b over two"
Next, let's move the small square with a side length of "b over two"

into the space above the lower rectangle
into the space above the lower rectangle

and subtract the lower rectangle from the two shapes above it.
and subtract the lower rectangle from the two shapes above it.

This leaves a square with a height and width of "x, minus b over two".
This leaves a square with a height and width of "x, minus b over two".

The area of this square created from the sum of these shapes
The area of this square created from the sum of these shapes

represents the quadratic expression "x-squared, minus bx, plus b over two, squared".
represents the quadratic expression "x-squared, minus bx, plus b over two, squared".

And since the side length of this square is "x, minus b over two"
And since the side length of this square is "x, minus b over two"

this quadratic expression can also be written as "x, minus b over two" quantity squared.
this quadratic expression can also be written as "x, minus b over two" quantity squared.

This geometric proof shows that when b is negative
This geometric proof shows that when b is negative

the quadratic can be written as "x, minus b over two" quantity squared.
the quadratic can be written as "x, minus b over two" quantity squared.

And we already know that when b is positive
And we already know that when b is positive

the quadratic can be written as "x, plus b over two" quantity squared.
the quadratic can be written as "x, plus b over two" quantity squared.

Therefore we can always use the same identity
Therefore we can always use the same identity

remembering that b can either be a positive or negative constant.
remembering that b can either be a positive or negative constant.

So let's try an example, solving a quadratic equation where b is negative.
So let's try an example, solving a quadratic equation where b is negative.

In this quadratic expression, the x-coefficient b is negative five
In this quadratic expression, the x-coefficient b is negative five

and the constant term c is three.
and the constant term c is three.

As we saw in "Part 1"
As we saw in "Part 1"

the first step is to move the constant term c to the right side
the first step is to move the constant term c to the right side

by subtracting it from both sides
by subtracting it from both sides

and then add "b over two, squared" to both sides.
and then add "b over two, squared" to both sides.

Since in this example, b is negative five
Since in this example, b is negative five

"b over two, squared" is "negative five over two" quantity squared
"b over two, squared" is "negative five over two" quantity squared

or squaring the numerator and denominator
or squaring the numerator and denominator

25 fourths.
25 fourths.

On the right side, we can simplify 25 fourths minus 3
On the right side, we can simplify 25 fourths minus 3

by converting three to 12 fourths
by converting three to 12 fourths

and then subtracting 12 fourths from 25 fourths
and then subtracting 12 fourths from 25 fourths

to get 13 fourths.
to get 13 fourths.

The quadratic expression on the left is now a perfect square
The quadratic expression on the left is now a perfect square

that as we saw, can be written as "x, plus b over two" quantity squared.
that as we saw, can be written as "x, plus b over two" quantity squared.

And since in this example, b is negative five
And since in this example, b is negative five

this perfect square quadratic can be written as "x minus five-halves" quantity squared.
this perfect square quadratic can be written as "x minus five-halves" quantity squared.

We then solve for x by taking the square root of both sides of the equation.
We then solve for x by taking the square root of both sides of the equation.

As we saw in "Part one", since the square root of the square of an expression
As we saw in "Part one", since the square root of the square of an expression

is the same as the absolute value of that expression
is the same as the absolute value of that expression

the square root and square can be replaced with the absolute value operation.
the square root and square can be replaced with the absolute value operation.

And since the absolute value of "x minus five-halves"
And since the absolute value of "x minus five-halves"

is equal to the square root of 13 fourths
is equal to the square root of 13 fourths

the value of "x minus five-halves "
the value of "x minus five-halves "

can be equal to either the positive or negative square root of 13 fourths.
can be equal to either the positive or negative square root of 13 fourths.

We can then move negative five-halves to the right side by subtracting it from both sides.
We can then move negative five-halves to the right side by subtracting it from both sides.

So the solution set of the quadratic equation "x-squared, minus five x, plus three, equals zero"
So the solution set of the quadratic equation "x-squared, minus five x, plus three, equals zero"

is "five-halves, plus-or-minus the square root of 13 fourths".
is "five-halves, plus-or-minus the square root of 13 fourths".

Likewise, the zeros of the function "x-squared, minus five x, plus three"
Likewise, the zeros of the function "x-squared, minus five x, plus three"

are "five-halves plus the square root of 13 fourths"
are "five-halves plus the square root of 13 fourths"

and "five-halves minus the square root of 13 fourths".
and "five-halves minus the square root of 13 fourths".

In the examples of quadratic equations that we have solved so far
In the examples of quadratic equations that we have solved so far

the constants b and c have been integers
the constants b and c have been integers

but as we have said, the completing the square method
but as we have said, the completing the square method

can be used to solve any quadratic equation.
can be used to solve any quadratic equation.

In our final example
In our final example

we will demonstrate that b and c don't have to be integers or even rational numbers
we will demonstrate that b and c don't have to be integers or even rational numbers

by solving the quadratic equation
by solving the quadratic equation

"x-squared, minus pi x, minus the square root of two, equals zero".
"x-squared, minus pi x, minus the square root of two, equals zero".

In this quadratic expression the x-coefficient b is "negative pi"
In this quadratic expression the x-coefficient b is "negative pi"

and the constant term c is "the negative square root of two".
and the constant term c is "the negative square root of two".

Once again, we start by moving the constant term c to the right side
Once again, we start by moving the constant term c to the right side

by subtracting it from both sides
by subtracting it from both sides

and then add "b over two, squared" to both sides.
and then add "b over two, squared" to both sides.

Then, to simplify these expressions a bit
Then, to simplify these expressions a bit

instead of squaring the entire fraction "negative pi over two"
instead of squaring the entire fraction "negative pi over two"

we can square the numerator and denominator separately.
we can square the numerator and denominator separately.

The quadratic expression on the left is now a perfect square
The quadratic expression on the left is now a perfect square

that can be written as "x, plus b over two" quantity squared.
that can be written as "x, plus b over two" quantity squared.

And since in this example, b is "negative pi"
And since in this example, b is "negative pi"

the perfect square quadratic can be written as "x, minus pi over two" quantity squared.
the perfect square quadratic can be written as "x, minus pi over two" quantity squared.

We can now solve for x by taking the square root of both sides of this equation.
We can now solve for x by taking the square root of both sides of this equation.

This formula gives us the two x values that are the solutions to the quadratic equation
This formula gives us the two x values that are the solutions to the quadratic equation

"x squared, minus pi x, minus the square root of two, equals zero".
"x squared, minus pi x, minus the square root of two, equals zero".

Since pi and the square root of two are irrational numbers
Since pi and the square root of two are irrational numbers

this formula represents two exact solutions
this formula represents two exact solutions

that could only be represented numerically by an infinite string of digits.
that could only be represented numerically by an infinite string of digits.

However, if we like, we can calculate approximate numerical values for these solutions
However, if we like, we can calculate approximate numerical values for these solutions

by filling in rounded values for pi
by filling in rounded values for pi

and the square root of two.
and the square root of two.

Completing the arithmetic
Completing the arithmetic

we see that the approximate solutions are 1.57 plus-or-minus 1.97 .
we see that the approximate solutions are 1.57 plus-or-minus 1.97 .

These two solutions could also be written separately
These two solutions could also be written separately

as 3.54
as 3.54

and negative 0.40.
and negative 0.40.

These are the zeros of the quadratic function
These are the zeros of the quadratic function

"x-squared, minus pi x, minus the square root of two".
"x-squared, minus pi x, minus the square root of two".

In the last few lectures, we have examined several methods of solving quadratic equations.
In the last few lectures, we have examined several methods of solving quadratic equations.

Before the method of completing the square was developed
Before the method of completing the square was developed

only very limited types of quadratic equations could be solved.
only very limited types of quadratic equations could be solved.

However, the technique of completing the square overcame this restriction.
However, the technique of completing the square overcame this restriction.

As mathematics progressed, this methodology was reduced to a formula.
As mathematics progressed, this methodology was reduced to a formula.

Given a general form quadratic equation
Given a general form quadratic equation

the constants a, b, and c could simply be plugged into this formula
the constants a, b, and c could simply be plugged into this formula

that then automatically gave the solutions to the equation.
that then automatically gave the solutions to the equation.

In the next lecture, we will introduce this "quadratic formula".
In the next lecture, we will introduce this "quadratic formula".