Download 0 = + + c bx ax 150 6 2 2 = + x 150 6 2 2 = + x 144 2 2 = x x . 72 x 72

Algebra II Pre-AP
Solving Quadratic Equations
IMPORTANT! PLEASE READ! -- For the benefit of students who are going to miss class because of the
band's trip to Indianapolis, I'm going to summarize the material that we will be covering while you are gone. Be
aware that NONE of this material is on Test #1C, which will take place on the Tuesday and Wednesday after
you return. ALL students -- including band students -- will be expected to take that test on time.
Consequently, students should prioritize preparation for that test and work through this material as time allows.
Please consider that I am trying to summarize at least 2 hours' worth of instruction here. Clearly, I can't give
you the same level of detail that I can give to students who are in class. In other words, these are the bullet
points only, so upon your return, you might want to get together with a classmate who is not in band to help fill
in any gaps.
Quadratic Equations are equations that can be written in the form ax 2  bx  c  0 , where a, b, and c are real
numbers and a  0 . You should probably ask yourself why a can't be 0. We are going to discuss 3 ways of
solve quadratic equations and in a future lesson we will summarize when it is best to use each method. Some
of the methods work all of the time while others do not. As you read through these notes, you should ask
yourself whether a method works always or only sometimes.
Please note that you already know how to solve quadratic equations by factoring, as this method was
discussed in relation to Assignment #20.
Method 1 -- The Square Root Method
Our goal is going to be to isolate either the x 2 or, failing that, an expression containing the x 2 . The other side
of the equation cannot contain any variable. We will then take the square root of both sides of the equation,
resulting in 2 linear equations that we can solve.
Example 1: Use the square root method to solve 2 x 2  6  150 .
2 x 2  6  150
2 x 2  144
x 2  72
x   72
Subtract 6 from both sides.
Don't take the square root yet! First divide by 2 to isolate the x 2 .
NOW take the square root of both sides.
Simplify the radical
x  6 2
Recall that we get 2 solutions because
x 2 is not x, nor is it  x , but rather
x 2  x . If you want to
convince yourself that this is true, graph y  x 2 on your calculator, and you will see the familiar v-shaped
graph. Really, then, the last half of Example 1 should look like this:
x 2  72
x 2  72
Take the square root of both sides
Replace
x 2 with x
x  72
Separate into 2 equations to get rid of the absolute value
x  72 or x   72
Simplify the radicals.
x  6 2 or x  6 2
You aren't expected to go through all of this every time, but I think that it is important to understand why you
usually get 2 solutions.
Example 2: Solve x 2  20  4 by using the square root method.
x 2  20  4
x 2  16
x    16
x  4i
Subtract 20 from both sides.
Take the square root of both sides.
Simplify the radical
Observe that in this example, the solutions are imaginary numbers. In Algebra I, you would have answered,
"No real solutions," but now that you know about imaginary numbers, you can specify the solutions.
2
n 
Example 3: Solve   1  16
3 
We really don't want to multiply out the left-hand side in hopes of
isolating the n 2 . An expression containing n 2 is already isolated,
so we can just take the square root of both sides.
n
 1   16
3
n
 1  4
3
n
n
 1  4 or
 1  4
3
3
n
n
 3 or  5
3
3
n  9 or n  15
Simplify the radical.
Separate this into 2 equations.
Subtract 1 from both sides of each equation
Multiply both sides of each equation by 3
It is fairly easy to substitute these values into the original equation to verify that they work.
Method 2 -- Completing the Square
When solving a quadratic equation by completing the square, your goal will be to get the equation to look like
the one in Example 3. Once that is accomplished, you can solve by taking the square root of both sides. Our
first challenge, then, is to get the equation into the proper form.
To understand how to do this, recall the pattern for squaring a binomial:  x  a   x 2  2a x  a 2 . I have put
2a in parentheses to emphasize that it is the coefficient of the linear x term. We are going to be doing this
process backwards, starting with the x 2  2a x and trying to figure out how to add a third term (in this case,
2
the a 2 ) so that we complete the trinomial so that it can be written as the square of a binomial. Let's see how
this works.
Example 4: Complete the square on n 2  12n .
We are trying to fill in the blanks of the following equation: n 2  12n  _____  ___  ___  . Our thinking
proceeds like this:
2
1) The second blank clearly just needs an "n." n 2  12n  _____  n  ___ 
2) The coefficient of the linear term on the left side is 12. We know that this is 2 times the number on
the third blank (this number on the third blank is what we referred to as "a" in the paragraph above.)
2
Since 2a  12 , the number on the third blank must be 6. n 2  12n  _____  n  6 .
2
3) The number on the first blank (previously referred to as " a 2 ") is just the square of the number on the
third blank. Thus, the number on the first blank must be 36. n 2  12n  36  n  6 . We have now
completed the square.
2
Example 5: Complete the square on k 2  3k
2
The solution is left to the student. You should get that k 2  3k 
9 
3
  k   . IMPORTANT NOTES: First
4 
2
observe that since the original binomial contained subtraction, the squared binomial also contains subtraction.
Second, notice that when b, the coefficient of the linear term, is odd, the squared binomial will contain a
fraction. Do NOT use decimals or mixed numbers! You will see why shortly.
Now we're ready to solve an equation by completing the square.
Example 6: Solve by completing the square: x 2  6 x  11  0
x 2  6 x  11  0
x 2  6 x  ___  11  ___
Move the constant to the right side and draw blanks on both sides.
Divide the 6 by 2 and write the left side as a squared binomial.
x  32  11  _____
Square the 3 and put the answer on ALL 3 blanks. The second
x  32  11  9
x  32  20
x  3   20
x  3  2 5
line should now read read: x 2  6 x  9  11  9 .
Combine like terms on the right.
Take the square root of both sides, remembering to use ±
Simplify the radical
Subtract 3 from both sides to isolate the variable.
x  3  2 5
Notice that, unlike Example 3, it was not necessary to split the equation into 2 separate equations after taking
the square root of both sides. This is because 2 5 is irrational and won't combine with the 3.
If the leading coefficient, a, is a number other than one, then you MUST begin by dividing everything on both
sides by that coefficient, as illustrated in Example 7 below.
Example 7: Solve by completing the square: 4 x 2  4 x  11  0 .
4 x 2  4 x  11  0
4 x 2 4 x 11

 0
4
4
4
11
x2  x   0
4
11
x 2  x  ___   ___
4
Divide everything by 4
Reduce if possible
Add the constant to both sides and insert blanks.
Complete the square by dividing the coefficient of x by 2.
2
1
11

 x     ___
2
4


x 

Square
1
and use the result to fill in all 3 blanks.
2
2
1
11 1
  
2
4 4
Combine like terms and reduce
2
1

x    3
2

1
x  3
2
1
x  3
2
Take the square root of both sides, remembering to use ±
Isolate the variable
Method 3 -- The Quadratic Formula
The quadratic formula expresses the solutions to the equation ax 2  bx  c  0 in terms of only a, b, and c. It
comes from completing the square on the equation ax 2  bx  c  0 . Doing so gives you the result that
 b  b 2  4ac
. You are probably familiar with this formula, so we'll just do one example and be done
x
2a
with this.
Example 8: Solve x 2  10 x  30 by using the quadratic formula.
x 2  10 x  30
x 2  10 x  30  0
a  1, b  10, c  30
Start by making one side zero
Identify the values of a, b, and c
Write the quadratic formula replacing a, b, and c with parentheses
 ( )  ( ) 2  4( )( )
x
2( )
Substitute values for a, b, and c
x
  10 
 102  4130
21
10  100  120
2
10   20
x
2
10  2i 5
x
2
10 2 5

x
i
2
2
x  5i 5
x
Do the multiplication
Do the subtraction
Simplify the radical
Since the answer has an imaginary part, split into 2 fractions
Reduce the fractions
Express the answer in a  bi form.
If the radicand had been positive instead of negative, then instead of splitting the answer into 2 fractions, you
would factor a 2 out of the numerator and reduce the fraction to x  5  5 .
I hope that this helps and that you have an enjoyable and successful trip!