Hello. I'm Professor Von Schmohawk and welcome to Why U.
In the previous lecture, we showed geometrically
that any quadratic expression of the form "x-squared, plus bx, plus b over two squared"
is a perfect square.
In this geometric proof
we represented the x-squared term
by the area of a square with sides of length x
the bx term
by the area of a rectangle with sides of lengths b and x
or alternatively, two smaller rectangles each with half that area
and the constant term "b over two" quantity squared
by the area of a square with sides of length "b over two".
The square created from the sum of these areas
therefore represented the entire quadratic expression.
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.
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
are added to the other areas rather than subtracted.
So what would this proof look like if the value of bx is negative?
Just as before, we can represent the x-squared term
as a square with sides of length x
the "b over two, squared" term
as a square with sides of length "b over two"
and the bx term
as two rectangles with sides of lengths "b over two" and x.
However, since the bx term is negative
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.
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"
into the space above the lower rectangle
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".
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".
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 geometric proof shows that when b is negative
the quadratic can be written as "x, minus b over two" quantity squared.
And we already know that when b is positive
the quadratic can be written as "x, plus b over two" quantity squared.
Therefore we can always use the same identity
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.
In this quadratic expression, the x-coefficient b is negative five
and the constant term c is three.
As we saw in "Part 1"
the first step is to move the constant term c to the right side
by subtracting it from both sides
and then add "b over two, squared" to both sides.
Since in this example, b is negative five
"b over two, squared" is "negative five over two" quantity squared
or squaring the numerator and denominator
25 fourths.
On the right side, we can simplify 25 fourths minus 3
by converting three to 12 fourths
and then subtracting 12 fourths from 25 fourths
to get 13 fourths.
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.
And since in this example, b is negative five
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.
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
the square root and square can be replaced with the absolute value operation.
And since the absolute value of "x minus five-halves"
is equal to the square root of 13 fourths
the value of "x minus five-halves "
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.
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".
Likewise, the zeros of the function "x-squared, minus five x, plus three"
are "five-halves plus 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
the constants b and c have been integers
but as we have said, the completing the square method
can be used to solve any quadratic equation.
In our final example
we will demonstrate that b and c don't have to be integers or even rational numbers
by solving the quadratic equation
"x-squared, minus pi x, minus the square root of two, equals zero".
In this quadratic expression the x-coefficient b is "negative pi"
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
by subtracting it from both sides
and then add "b over two, squared" to both sides.
Then, to simplify these expressions a bit
instead of squaring the entire fraction "negative pi over two"
we can square the numerator and denominator separately.
The quadratic expression on the left is now a perfect square
that can be written as "x, plus b over two" quantity squared.
And since in this example, b is "negative pi"
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.
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".
Since pi and the square root of two are irrational numbers
this formula represents two exact solutions
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
by filling in rounded values for pi
and the square root of two.
Completing the arithmetic
we see that the approximate solutions are 1.57 plus-or-minus 1.97 .
These two solutions could also be written separately
as 3.54
and negative 0.40.
These are the zeros of the quadratic function
"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.
Before the method of completing the square was developed
only very limited types of quadratic equations could be solved.
However, the technique of completing the square overcame this restriction.
As mathematics progressed, this methodology was reduced to a formula.
Given a general form quadratic equation
the constants a, b, and c could simply be plugged into this formula
that then automatically gave the solutions to the equation.
In the next lecture, we will introduce this "quadratic formula".