7. Solving Quadratic Equations by Completing the Square
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Chapter 8
7.
Solving Quadratic Equations by Completing the Square
Completing the square is a technique used to solve quadratic equations. The method involves changing the equation into a perfect square trinomial, simplifying the process of finding solutions. Real-world applications include calculating pool dimensions, determining rocket heights, and figuring out how long it takes for a lemonade dispenser to empty. This approach is versatile and can tackle equations that other methods may find challenging. It is especially useful when the equation is not easily factorable or when precise numerical solutions are needed.
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| | 9 Theory slides |
| | 9 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Solving Quadratic Equations by Completing the Square
Slide of 9
In perfect square trinomials with leading coefficient 1, there is a relationship between the coefficient of the x-term and the constant.
x2+bx+(2b)2
Considering this relationship, quadratic expressions in the form x2+bx can be transformed into perfect square trinomials. This lesson will present how these expressions can be transformed into perfect square trinomials and their applications for solving quadratic equations. Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Challenge
Is It Possible to Rewrite Equations That Are in Standard Form to Find Their Solutions?
Paulina's birthday is this weekend and her parents have hidden her gifts in a trunk. She can have them early if she can open the combination lock on the truck. Her parents gave her the clues to help her find the combination, and she has figured out all but the final two digits.
x2+2x−24=0
While considering the equation, Paulina wonders if there is a way to rewrite the given equation so that the variable term appears only once on the left-hand side. If this is possible, what would be its advantages?Discussion
Completing the Square
In a perfect square trinomial, there is a relationship between the coefficient of the x-term and the constant term — the constant term is equal to the square of half the coefficient of the x-term.
This process is called completing the square. To complete the square for an expression algebraically, these steps can be followed.
x2+bx+(2b)2
This relationship can be used to form a perfect square trinomial by adding a constant c to any expression in the form x2+bx.
x2+bx⇒x2+bx+c
The process of finding the constant c can be visualized by using algebraic tiles. Consider the following expression. x2+6x
The expression is represented using algebraic tiles. Then, a square is created by rearranging the existing tiles and adding more tiles. The following applet summarizes this process.
1
Identify the Coefficient of the x-Term
expand_more For the given expression, the value of b is 6.
x2+6x
2
Calculate (2b)2
expand_more Once the value of b is identified, calculate the square of half of the value of b.
3
Add (2b)2 to the Initial Expression
expand_more Add (2b)2 to the expression to obtain a perfect square trinomial.
x2+bx⇒x2+bx+(2b)2
In this case, 9 should be added to x2+6x.
x2+6x⇒x2+6x+9
4
Factor the Perfect Square Trinomial
expand_more The expression obtained in the previous step can be now factored as the square of a binomial.
Therefore, a perfect square trinomial is obtained by adding a constant c=(2b)2 to the initial expression in the form x2+bx.
x2+bx+(2b)2=(x+2b)2
This will be applied to the expression x2+6x+9.
x2+6x+9
SplitIntoFactors
Split into factors
x2+2⋅3x+9
CommutativePropMult
Commutative Property of Multiplication
x2+2x⋅3+9
FacPosPerfectSquare
a2+2ab+b2=(a+b)2
(x+3)2
Pop Quiz
Practicing Finding the Values for Completing the Square
In the following applet, use the method of completing the square to determine the value of c that makes the given expression a perfect square trinomial. Round to 2 decimal places if needed.
Discussion
Extending Completing the Square to Quadratic Equations
The most useful application of completing the square is that it can be extended to solve quadratic equations. However, some additional steps need to be taken when using this method to solve equations.
Method
Solving a Quadratic Equation by Completing the Square
To solve a quadratic equation by completing the square, write the equation in the form x2+bx=c, then complete the square of the expression on the left-hand side. Finally, the equation can be solved for x by taking square roots of both sides of the equation. To illustrate this, consider the following equation.
2x2+12x+4=0
To solve the equation by completing the square, these five steps can be followed.
1
Write the Equation in the Form x2+bx=c
expand_more The Properties of Equality can be used jointly with inverse operations to rewrite the given equation in the form x2+bx=c.
If the equation is already in the form x2+bx=c, this step is skipped.
2x2+12x+4=0
SplitIntoFactors
Split into factors
2x2+2⋅6x+2⋅2=0
FactorOut
Factor out 2
2(x2+6x+2)=0
DivEqn
LHS/2=RHS/2
x2+6x+2=0
SubEqn
LHS−2=RHS−2
x2+6x=-2
2
Complete the Square on the Left-Hand Side of the Equation
expand_more To complete the square on the left-hand side of the equation, the square of one-half the coefficient of the x-term should be added to each side of the equation.
x2+bx=c⇓x2+bx+(2b)2=c+(2b)2
In the equation found previously, b is equal to 6. Therefore, (26)2 should be added to both sides.
x2+6x=-2⇓x2+6x+(26)2=-2+(26)2
Next, the quotient can be simplified.
x2+6x+32=-2+32
The expression on the left-hand side is now a perfect square trinomial. 3
Factor the Perfect Square Trinomial
expand_more Next, factor the perfect square trinomial.
x2+6x+32=-2+32
SplitIntoFactors
Split into factors
x2+2⋅x⋅3+32=-2+32
FacPosPerfectSquare
a2+2ab+b2=(a+b)2
(x+3)2=-2+32
(x+3)2=7
4
Take the Square Root of Both Sides of the Equation
expand_more The square root of both sides of the equation can now be taken to remove the exponent.
5
Solve for x
expand_more Finally, the resulting equations of the previous step need to be solved. These solutions will also be solutions to the original equation.
| x+3=±7 | ||
|---|---|---|
| Write as two equations | x+3=7 | x+3=-7 |
| Solve for x | x=-3+7 | x=-3−7 |
Therefore, the solutions of the given equation are x=-3+7 and x=-3−7.
Example
Solving Quadratic Equations by Completing the Square
While Paulina thinks about finding the missing digits of the combination lock, her older brother, Vincenzo, and her parents are setting up a rectangular pool for her birthday party. The pool will be in the backyard and will cover an area of 768 square feet. Additionally, they want the length of the pool to be 32 feet longer than the width.
External credits: @kdekiara
Answer the following questions to help Vincenzo and his parents find the dimensions they should use for the pool.
a Write a quadratic equation representing the area of the pool in terms of the width x.
b Find and interpret the solutions of the quadratic equation by completing the square.
Answer
a x2+32x=768
b Solutions: x=16 and x=-48
Interpretation: The dimensions of the pool cannot be negative. Therefore, the pool will have a width of 16 feet and a lenght of 48 feet.
Interpretation: The dimensions of the pool cannot be negative. Therefore, the pool will have a width of 16 feet and a lenght of 48 feet.
Hint
a Use the formula for the area of a rectangle.
b The dimensions of the pool cannot be negative.
Solution
a The formula for the area of a rectangle will be used to create the quadratic equation representing the area of the pool.
A=ℓw
In this formula, A represents the area, ℓ the length, and w the width of the rectangle. It is given that the area of the pool must be 768 square feet.
A=ℓwSubstitute768=ℓw
Now, let x represent the width of the pool. Because the length will be 32 feet longer than the width, it can be represented as x+32. 
External credits: @kdekiara
b To solve the quadratic equation written in Part A, the method of completing square will be used. Notice that the equation is already in the form x2+bx=c.
x2+32x=768
To complete the square on the left-hand side of the equation, (2b)2 needs to be calculated. In this case, b=32.
(232)2⇔162
This term needs to be added to both sides of the equation to produce an equivalent equation. x2+32x+162=768+162
The perfect square trinomial can now be factored. x2+32x+162=768+162
SplitIntoFactors
Split into factors
x2+2⋅x⋅16+162=768+162
FacPosPerfectSquare
a2+2ab+b2=(a+b)2
(x+16)2=768+162
CalcPow
Calculate power
(x+16)2=768+256
AddTerms
Add terms
(x+16)2=1024
| x+16=±32 | |
|---|---|
| x+16=32 | x+16=-32 |
| x=16 | x=-48 |
Width: Length: 16 ft216+32=48 ft2
Example
Completing the Square to Solve Daily Problems
At Paulina's birthday party, there will be a lemonade dispenser that automatically fills people's glasses. The dispenser has a capacity of 40 liters and it is expected to be emptied after 60 minutes. The dispenser must be refilled when there is only one liter of lemonade left in it in order for the automatic filling function to work.
V=t2−110t+3026
Use the method for completing the square to find how long it will take to have only one liter left in the dispenser. {"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":true,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":"minutes","answer":{"text":["55"]}}
Hint
Substitute 1 into the equation for V and solve it by completing the square.
Solution
It is asked to find how long it will take to have only one liter left in the dispenser. To do so, substitute 1 for V and solve the equation by completing the square.
Now, the square root can be applied to both sides of the equation to find its solutions. However, because the right-hand side of the equation is 0, there will be only one solution for the equation.
This means that there is 1 liter left in the dispenser after 55 minutes.
1=t2−110t+3026⇕t2−110t+3026=1
The first step is to write the equation in the form x+bx=c.
For this equation, the value of b is -110. With this information, (2b)2 can be calculated.
(2-110)2⇔(-55)2=552
A perfect square trinomial is formed on the left-hand side of the equation by adding this term to both sides of the equation. This trinomial can then be factored.
t2−110t=-3025
AddEqn
LHS+552=RHS+552
t2−110t+552=-3025+552
SplitIntoFactors
Split into factors
t2−2⋅t⋅55+552=-3025+552
FacNegPerfectSquare
a2−2ab+b2=(a−b)2
(t−55)2=-3025+552
(t−55)2=0
(t−55)2=0
▼
Solve for t
SqrtEqn
LHS=RHS
(t−55)2=0
CalcRoot
Calculate root
(t−55)2=0
SqrtPowToNumber
a2=a
t−55=0
AddEqn
LHS+55=RHS+55
t=55
Example
Completing the Square to Identify Quadratic Equations With No Solutions
Dominika and Heichi built a small rocket for Paulina's birthday party. They all excitedly decide to launch it at the end of her birthday party.
h=-16t2+32t+12
The friends are fascinated by the upcoming launch of the rocket. Now they would like to discover if the rocket will reach a height of 60 feet above the ground. Complete the square to help them discover if the rocket can reach this height. Interpret the solution.
Answer
No, see solution.
Hint
The square of a real number cannot be negative.
Solution
To find if the rocket can reach a height of 60 feet, this value will be substituted into the given equation for h.
Notice that the value of b in this equation is -2. To complete the square on the left-hand side, the square of half the value of b, (2-2)2, should be added to both sides.
Note that the right-hand side is negative. Since there is no real number that has a negative square, the equation has no real solutions. This means that the rocket cannot reach a height of 60 feet.
60=-16t2+32t+12⇕-16t2+32t+12=60
The equation must first be written in the form x2+bx=c to solve it by completing the square. -16t2+32t+12=60
SubEqn
LHS−12=RHS−12
-16t2+32t=48
▼
Rewrite
MultEqn
LHS⋅(-1)=RHS⋅(-1)
(-16t2+32t)(-1)=48(-1)
Distr
Distribute (-1)
(-16t2)(-1)+(32t)(-1)=48(-1)
MultNegNegTwoPar
(-a)(-b)=a⋅b
16t2+(32t)(-1)=48(-1)
MultPosNeg
a(-b)=-a⋅b
16t2+(-32t)=-48
AddNeg
a+(-b)=a−b
16t2−32t=-48
SplitIntoFactors
Split into factors
16⋅t2−16⋅2⋅t=-16⋅3
FactorOut
Factor out 16
16(t2−2t)=-16⋅3
DivEqn
LHS/16=RHS/16
t2−2t=16-16⋅3
MoveNegNumToFrac
Put minus sign in front of fraction
t2−2t=-1616⋅3
DenomMultFracToNumber
A⋅Aa=a
t2−2t=-3
(2-2)2⇔(-1)2=1
Therefore, adding 1 to both sides of the equation will produce a perfect square trinomial on the left-hand side that can be factored as the square of a binomial.
t2−2t=-3
AddEqn
LHS+1=RHS+1
t2−2t+1=-3+1
FacNegPerfectSquare
a2−2ab+b2=(a−b)2
(t−1)2=-3+1
AddTerms
Add terms
(t−1)2=-2
Closure
Completing the Square to Decipher the Code
In this lesson, quadratic expressions in the form x2+bx were turned into perfect square trinomials in a process called completing the square. Similarly, quadratic equations were solved by completing the square. 
She remembers that the last digit 2 units greater than the one before it. Also, the second to last digit is a solution to the following quadratic equation.
x2+bx=c⇓Perfect Square Trinomialx2+xb+(2b)2=c+(2b)2
Considering these methods, the challenge presented at the beginning can now be solved. Recall that Paulina's parents gave her the clues to the combination of the lock on the trunk containing her birthday gifts. Paulina has figured out all but the last two digits of the combination. x2+2x−24=0
The given equation can be rewritten by completing the square. The advantage of completing the square is that every equation can be solved using this method. Also, once the square is completed, determining the solutions is straightforward. Using the given information, answer the following questions to help Paulina. a What is the correct number for the second to last digit?
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b What is the last digit of the combination?
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Hint
a Consider the the lock contains only non-negative numbers.
b Add 2 to the second to last number of the code.
Solution
a To find the second to last digit of the code, the given equation will be solved by completing the square. The equation will first be written in the form x+bx=c.
x2+2x−24=0⇓x2+2x=24
In this equation, b equals 2. With this information, the missing term (2b)2 can be determined.
(22)2⇔1
Adding 1 to both sides of the equation will produce a perfect square trinomial on the left-hand side.
Now, take the square root of both sides of the equation.
This can be written as two equations and by solving these equations, the solutions of the quadratic equation are obtained. | x+1=±5 | |
|---|---|
| x+1=5 | x+1=-5 |
| x=4 | x=-6 |
There are two solutions for the equation. However, since the combination lock does not include negative numbers, only 4 makes sense. Therefore, the second to last number of the code is 4.
b Paulina knows that the last number is 2 units greater than the number before it. Therefore, the last number is 4+2=6. Now she can fully decipher the code and open the wooden trunk to discover her birthday gifts.
Which expression does not belong to the group?
A.B.C.D.m2+23m+169m2−3m+49m2+4m+4m2+32m+92
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Observing the expressions, it seems they are perfect square trinomial. To verify this, we will calculate the square of half of the value of the coefficient of the linear term, ( b2)^2. Let's calculate it for the first expression. m^2+ 3/2m+9/16 In this case, b= 32.
Following a similar fashion, we can calculate ( b2)^2 for the remaining expressions.
| Expression | (b/2)^2 | |
|---|---|---|
| A | m^2+3/2m+ 9/16 | (32/2)^2= 9/16 |
| B | m^2-3m+ 9/4 | (-3/2)^2= 9/4 |
| C | m^2+4m+ 4 | (4/2)^2= 4 |
| D | m^2+2/3m+2/9 | (23/2)^2=1/9 |
From the table, we can see that for the cases of A, B, and C, the term ( b2)^2 is equal to the constant term c. This means that expressions A, B, and C can all be written as perfect squares and D cannot. Therefore, expression D does not belong to the group.
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The sides of a rectangle are x−1 and x+2.
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The area of a rectangle is the product of its length l and its width w. A=l w In this case, the given rectangle has a width of x-1 and its length is x+2. Substituting these expressions into the formula will give us an expression for the area of the given rectangle.
We are told that the area of the rectangle is 13.75 square centimeters. We will substitute 13.75 for A into the equation found previously.
Solving this equation will give us the possible values for x. Note that the expression is already in the form x^2+bx=c. This means we can solve the equation by completing the square. To do so, we need to add ( b2)^2 to both sides of the equation. In this case, b= 1. b= 1 ⇒ (b/2)^2=(1/2)^2 Let's now add this value to both sides of the equation.
We can now take the square root to both sides of the equation to obtain two linear equations whose solutions are also solutions to the quadratic equation.
Using the positive and negative signs will give us the possible values for x.
| x=-0.5±4 | ||
|---|---|---|
| x=-0.5+4 | x=-0.5-4 | |
| x=3.5 | x=-4.5 | |
We have found the possible values for x. Since a length cannot be negative, only x=3.5 makes sense in this context. Therefore, the value of x is 3.5 centimeters.
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A farmer is putting up fencing around a rectangular area for his garden. He plans to enclose the garden using fencing on three sides. The farmer has budgeted enough money for 67 ft of fencing material and would like to make a garden with an area of 560 ft2.
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The farmer plans to enclose the garden using fencing on three sides. Let w be the width and l the length of the garden. We know that the budget is enough to buy 67ft of fencing. This means 67 should equal two times w plus l. 2w+l= 67 We will isolate l to get an expression for the length of the garden in terms of w. 2w+l=67 ⇔ l=67-2w
We are told that the area of the garden should equal 560ft^2. Consider that the area of a rectangle is equal to its width times its length. This means we need to equate the product of w and l to 560. 560=w*l Previously, we have found that l=67-2w. We can substitute this expression into the equation for the area of the garden and then solve for w.
We have found a quadratic equation and we will now solve it by completing the square. Let's divide both sides of the equation by -2 to write it in the form x^2+bx=c.
In a quadratic expression, b is the coefficient of the linear term. We will calculate the missing term ( b2 )^2 needed to complete the square. In this case, b=- 672.
Next, we will add this term to both sides of our equation. Then we will factor the resulting perfect square trinomial.
The solutions for this equation are w= 674± 34. Let's separate them into the positive and negative cases.
| w= 674± 34 | |
|---|---|
| w_1=67/4+ 3/4 | w_2=67/4- 3/4 |
| w_1=17.5 | w_2=16 |
We have two solutions to the equation. The options for the width of the garden are 17.5ft or 16ft.
We have found that the length of the garden is given by l=67-2w. Additionally, the options for the width are 17.5ft and 16ft. Let's substitute these values into the expression for the length to find the possible lengths for the garden.
| w | 67-2w | l= 67-2w |
|---|---|---|
| w_1= 17.5 | 67-2* 17.5 | l=32 |
| w_2= 16 | 67- 2* 16 | l=35 |
The length of the garden will be 32ft if the width is 17.5ft and 35ft if the width is 16ft.
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Solving Quadratic Equations by Completing the Square
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