10.4: Solving Quadratic Equations by Completing the Square
There are several ways to write an equation for a parabola:
 Standard form:
y=ax2+bx+c  Factored form:
y=(x+m)(x+n) 
Vertex form:
y=a(x−h)2+k
Vertex form of a quadratic equation:
Example 1: Determine the vertex of
Solution: Using the definition of vertex form,
 The vertex is (4, –7).
 Because
a is negative, the parabola opens downward.  Therefore, the vertex (4, –7) is a maximum point of the parabola.
Once you know the vertex, you can use symmetry to graph the parabola.



2  
3  
4  –7 
5  
6 
Example 2: Write the equation for a parabola with
Solution: Using the definition of vertex form
y &= 3(x+4)^2+5
Consider the quadratic equation
Completing the Square
Completing the square is a method used to create a perfect square trinomial, as you learned in the previous chapter.
A perfect square trinomial has the form
Example: Find the missing value to create a perfect square trinomial:
Solution: The value of
a \ \text{is} \ x, && 2(xb) &= 8x\\
\text{Solve for} \ b: && \frac{2xb}{2x} &= \frac{8x}{2x} \rightarrow b=4
To complete the square you need the value of
The missing value is 16.
To complete the square, the equation must be in the form:
Looking at the above example,
Example 3: Find the missing value to complete the square of
Solution: Use the definition of the middle term to complete the square.
Therefore,
Solve Using Completing the Square
Once you have the equation written in vertex form, you can solve using the method learned in the last lesson.
Example: Solve
Solution: By completing the square and factoring, the equation becomes:
\text{Solve by taking the square root:} && x+11 &= \pm0\\
\text{Separate into two equations:} && x+11 &=0 \ or \ x+11=0\\
\text{Solve for} \ x: && x &= 11
Example: Solve
Solution: Using the definition to complete the square,
Therefore, to complete the square, we must rewrite the standard form of this equation into vertex form.
Subtract 9:
Complete the square: Remember to use the Addition Property of Equality.
\text{Factor the left side.} && (x+5)^2 &= 16\\
\text{Solve using square roots.} && \sqrt{(x+5)^2} &= \pm \sqrt{16}\\
&& x+5 &=4 \ or \ x+5=4\\
&& x &= 1 \ or \ x=9
Example: An arrow is shot straight up from a height of 2 meters with a velocity of 50 m/s. What is the maximum height that the arrow will reach and at what time will that happen?
Solution: The maximum height is the vertex of the parabola. Therefore, we need to rewrite the equation in vertex form.
&& y2 &= 4.9t^2+50t\\
&& y2 &= 4.9(t^210.2t)\\
\text{Complete the square inside the parentheses.} && y24.9(5.1)^2 &= 4.9(t^210.2t+(5.1)^2)\\
&& y129.45 &= 4.9(t5.1)^2
The maximum height is 129.45 meters.
Multimedia Link: Visit the http://www.mathsisfun.com/algebra/completingsquare.html  mathisfun webpage for more explanation on completing the square.
Practice Set
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both.
CK12 Basic Algebra: Solving Quadratic Equations by Completing the Square (14:06)
 What does it mean to “complete the square”?
 Describe the process used to solve a quadratic equation by completing the square.
 Using the equation from the arrow in the lesson,
 How high will an arrow be four seconds after being shot? After eight seconds?
 At what time will the arrow hit the ground again?
Write the equation for the parabola with the given information.

a=a , vertex=(h,k) 
a=13 , vertex=(1,1) 
a=−2 , vertex=(−5,0)  Containing (5, 2) and vertex (1, –2)

a=1 , vertex=(−3,6)
Complete the square for each expression.

x2+5x 
x2−2x 
x2+3x 
x2−4x 
3x2+18x 
2x2−22x 
8x2−10x 
5x2+12x
Solve each quadratic equation by completing the square.

x2−4x=5 
x2−5x=10 
x2+10x+15=0 
x2+15x+20=0 
2x2−18x=0 
4x2+5x=−1 
10x2−30x−8=0 
5x2+15x−40=0
Rewrite each quadratic function in vertex form.

y=x2−6x 
y+1=−2x2−x 
y=9x2+3x−10 
y=32x2+60x+10
For each parabola, find:
 The vertex

x− intercepts 
y− intercept  If it opens up or down
 The graph the parabola

y−4=x2+8x 
y=−4x2+20x−24 
y=3x2+15x 
y+6=−x2+x 
x2−10x+25=9 
x2+18x+81=1 
4x2−12x+9=16 
x2+14x+49=3 
4x2−20x+25=9 
x2+8x+16=25  Sam throws an egg straight down from a height of 25 feet. The initial velocity of the egg is 16 ft/sec. How long does it take the egg to reach the ground?
 Amanda and Dolvin leave their house at the same time. Amanda walks south and Dolvin bikes east. Half an hour later they are 5.5 miles away from each other and Dolvin has covered three miles more than the distance that Amanda covered. How far did Amanda walk and how far did Dolvin bike?
 Two cars leave an intersection. One car travels north; the other travels east. When the car traveling north had gone 30 miles, the distance between the cars was 10 miles more than twice the distance traveled by the car heading east. Find the distance between the cars at that time.
Mixed Review
 A ball dropped from a height of four feet bounces 70% of its previous height. Write the first five terms of this sequence. How high will the ball reach on its
8th bounce?  Rewrite in standard form:
y=27x−11 .  Graph
y=5(12)x . Is this exponential growth or decay? What is the growth factor?  Solve for
r:3r−4≤2 .  Solve for
m:−2m+6=−8(5m+4) .  Factor
4a2+36a−40 .
Quick Quiz
 Graph
y=−3x2−12x−13 and identify: The vertex
 The axis of symmetry
 The domain and range
 The
y− intercept  The
x− intercepts estimated to the nearest tenth
 Solve
y=x2+9x+20 by graphing.  Solve for
x:74=x2−7 .  A baseball is thrown from an initial height of 5 feet with an initial velocity of 100 ft/sec.
 What is the maximum height of the ball?
 When will the ball reach the ground?
 When is the ball 90 feet in the air?
 Solve by completing the square:
v2−20v+25=6