Hello. I'm Professor Von Schmohawk and welcome to Why U.
So far, we have seen that quadratic equations can be solved
if we can factor the quadratic expression into a pair of linear expressions.
However, there are many quadratics that are not practical or even possible to factor
using any of the methods we have presented so far.
Over one thousand years ago problems like this prompted mathematicians
to search for a technique that could be used to solve any quadratic equation.
The break-through in devising a general method for solving quadratic equations
was the development of a technique called "Completing the Square"
by the Persian astronomer and mathematician al-Khwarizmi.
The basic idea behind completing the square
is to transform a quadratic equation
containing a quadratic expression that is difficult to factor
into an equation with identical solutions
but containing a quadratic expression that is a perfect square.
Since this transformation also changes the right side of the equation
the perfect square quadratic expression
ends up being equated to a non-zero constant instead of zero
so the "zero product property" cannot be used.
However, since the quadratic expression is a perfect square
it can now be rewritten as the square of a linear expression.
The square root of both sides of the equation can then be taken
to simplify the equation into a form that can be easily solved.
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+3 is equal to the square root of 2
the value of x+3 can be equal to either the positive or negative square root of 2.
To solve for x we then move 3 to the right side
by subtracting 3 from both sides
giving us the values of x that are the solutions to the quadratic equation.
To understand the concept of completing the square
let's start by comparing two quadratic functions
"x-squared plus 6x plus 7"
and "x-squared plus 6x plus 9".
These two functions are identical
except for their constant terms 7 and 9.
Since 9 is two greater than the 7
the right graph is identical
but shifted two units higher.
However, there is another important difference.
The function on the right is a perfect square.
Since this quadratic expression is a perfect square
it has two identical linear factors.
In this example the two factors are both x+3
so this perfect square quadratic can also be written as "x+3 quantity squared".
In fact, all perfect square quadratics
can be written as the square of a single linear expression
which is why they are called "perfect squares".
Since identical linear factors share the same zero
perfect square quadratic functions have only a single x-intercept
and so the vertexes of their parabolic graphs must lie on the x-axis.
On the other hand, the graph on the left has two x-intercepts
so that quadratic function is not a perfect square.
If we set this function equal to zero
the solutions to this quadratic equation will be the function's x-intercepts.
But how can we determine these values of x?
This quadratic expression cannot be factored
using any of the techniques we have learned so far
so we can't find the solutions by factoring
and then using the "zero product property".
However, by completing the square
the solutions to this equation can be found.
If we add 2 to each side of this equation
the quadratic's constant term is increased by 2
shifting the parabola up by two units.
This quadratic function is now the same as the function on the right.
But since the right side of the quadratic equation has also been increased by 2
the solutions to the equation
are no longer the values of x where the function is equal to zero.
They are now the values of x where the function is equal to 2.
These are the same x values where the original function was equal to zero.
Therefore, the solutions to this equation will be the same x values
as the solutions to the original equation.
But unlike the original equation we now have a way to solve this equation.
Since this quadratic expression is a perfect square
it can be written as the square of a single linear expression
and as we saw, in this case the squared linear expression is x+3.
This equation can now be solved by taking the square root of both sides.
And as we saw
since one side of the equation is the square root of a squared expression
the square root of the square can be eliminated
as long as a plus-or-minus sign is added to the opposite side of the equation.
We can then solve for x by subtracting 3 from both sides.
This gives us the two x values that are the solutions to the quadratic equation.
So we have shown that by changing the value of a quadratic expression's constant term
we can change any quadratic expression into a perfect square.
But how can we determine what the value of this constant term should be?
In the example we just saw
we knew that a constant term of 9 would produce a perfect square
but normally we don't know this value and so this constant must be calculated.
Fortunately, this is an easy calculation
and we can illustrate graphically how this value is determined.
Given any general form quadratic equation
the first step is to eliminate the x-squared coefficient.
This can always be accomplished
by dividing both sides of the equation by that coefficient.
For example, to eliminate the x-squared coefficient
in the equation "4 x-squared, plus 3x, plus 8 equals zero"
we divide both sides of the equation by 4.
"4 divided by 4" is one.
"8 divided by 4" is 2.
And "zero divided by 4" is zero.
The x-squared coefficient of "one" can now been eliminated.
And since both sides of the equation were divided by the same number
the solutions to the equation have not been affected.
Therefore in this lecture, we will always start with a quadratic equation
that has been put into a form where the x-squared coefficient is "one".
So, how do we determine the value of a constant term
that will produce a perfect square?
We will show that if the constant term of a quadratic expression in this form
has a value of "b over two" quantity squared
that quadratic expression will be a perfect square.
We can show why this is true
by representing the quadratic by a group of squares and rectangles.
We will represent the x-squared term
by a square with sides of length x.
The area of that square is therefore x-squared.
Likewise, we will represent the term "bx"
by a rectangle with sides of length b and x
and thus an area of bx
or alternatively, two smaller rectangles, each with half that area.
Then, all we need to "complete the square"
is a small square with sides of length "b over two"
and thus an area of "b over two" squared.
So if we make the value of a quadratic's constant term "b over two" squared
then that quadratic will be a perfect square.
Since the sides of this square have a length of x plus "b over two"
the area of the square can be written as either "x, plus b over two" quantity squared
or the quadratic expression x-squared, plus bx, plus "b over two" squared.
So if a quadratic with an x-squared coefficient of "one"
has a constant term equal to "b over two" squared
then that quadratic expression will be a perfect square
and can also be written as "x, plus b over two" quantity squared.
Using this fact, we can solve any quadratic equation
in the form "x-squared, plus bx, plus c equals zero"
by changing the quadratic's constant term to have a value of "b over two" squared.
This can always be accomplished by subtracting c from both sides
and then adding "b over two" squared to both sides.
This gives us a perfect square expression
equal to a constant.
We can then write the perfect square quadratic as "x, plus b over two" quantity squared.
The values of x can then be found by taking the square root of both sides of the equation.
So let's try an example, using this technique to solve the quadratic equation
"x-squared plus 2x minus 2" equals zero.
These solutions can be found by completing the square.
In this quadratic expression the x-coefficient b is 2
and the constant term c is negative 2.
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 2
"b over two" squared, is "two over two" squared
or "one".
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 2
this perfect square quadratic can be written as x+1 quantity squared.
Once again, we solve for x by taking the square root of both sides of the equation.
So the solution set of the quadratic equation "x-squared plus 2x minus 2" equals zero
is "negative-one plus-or-minus the square root of 3".
Likewise, the zeros of the function "x-squared plus 2x minus 2"
are "negative-one plus the square root of 3"
and "negative-one minus the square root of 3".
In the lecture "Factoring Quadratic by Inspection - part 1"
we said that the quadratic function "x-squared plus 9x plus 12"
was not able to be factored by inspection.
But as we saw in that lecture the graph of this function
crosses the x-axis at the two points
x equals negative 9 plus-or-minus the square root of "33 over 2".
So let's confirm this by setting this function equal to zero
and solving that quadratic equation by completing the square.
In this quadratic expression the x-coefficient b, is 9
and the constant term c, is 12.
Just as before, we 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.
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 nine
this perfect square quadratic can be written as "x plus nine-halves" quantity squared.
We can then simplify the right side of the equation
by squaring the fraction nine-halves
squaring its numerator and denominator.
81 fourths and 12 can then be combined
by first writing 12 as 48 fourths
and then subtracting 48 fourths from 81 fourths
giving us 33 fourths.
We then solve for x by taking the square roots of both sides of the equation.
We can simplify further by writing 33 fourths
as one-fourth times 33.
And since one-fourth
is one-half squared
it can be brought out from under the radical sign.
Finally, negative 9 halves plus-or-minus the square root of 33 halves
can be combined into a single fraction
"negative-9 plus-or-minus the square root of 33" over 2.
So although the quadratic function "x-squared plus 9x plus 12"
was not able to be factored by inspection
the zeros of that function can be calculated by completing the square.
We have shown geometrically that any quadratic expression of the form
x-squared, plus bx, plus "b over two" squared is a perfect square.
In the next lecture, we will show what happens to this geometric proof
when the bx term is negative.