Sign In
| 9 Theory slides |
| 9 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
x^2+bx+(b/2 )^2 Considering this relationship, quadratic expressions in the form x^2+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.
Here are a few recommended readings before getting started with this lesson.
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.
Paulina knows that the last digit is 2 units greater than the one before it. She also knows that the second to last digit is a solution to the following quadratic equation. x^2+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?For the given expression, the value of b is 6. x^2+ 6x
Add ( b2)^2 to the expression to obtain a perfect square trinomial. x^2+bx ⇒ x^2+bx+ (b/2)^2 In this case, 9 should be added to x^2+6x. x^2+6x ⇒ x^2+6x+ 9
Split into factors
Commutative Property of Multiplication
a^2+2ab+b^2=(a+b)^2
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.
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.
Split into factors
Factor out 2
.LHS /2.=.RHS /2.
LHS-2=RHS-2
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. x^2 + bx = c ⇓ x^2 + bx +(b/2)^2 = c +(b/2)^2 In the equation found previously, b is equal to 6. Therefore, ( 62)^2 should be added to both sides. x^2 + 6x = - 2 ⇓ x^2 + 6x +(6/2)^2 = - 2 +(6/2)^2 Next, the quotient can be simplified. x^2+6x+3^2=-2+3^2 The expression on the left-hand side is now a perfect square trinomial.
Split into factors
a^2+2ab+b^2=(a+b)^2
Finally, the resulting equations of the previous step need to be solved. These solutions will also be solutions to the original equation.
x+3 = ± sqrt(7) | ||
---|---|---|
Write as two equations | x+3 = sqrt(7) | x+3 = - sqrt(7) |
Solve for x | x =- 3+ sqrt(7) | x = -3- sqrt(7) |
Therefore, the solutions of the given equation are x=- 3+sqrt(7) and x=- 3-sqrt(7).
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.
Answer the following questions to help Vincenzo and his parents find the dimensions they should use for the pool.
A=l w 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=l w Substitute 768=l w 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.
Split into factors
a^2+2ab+b^2=(a+b)^2
Calculate power
Add terms
sqrt(LHS)=sqrt(RHS)
sqrt(a^2)=± a
Calculate root
x+16= ± 32 | |
---|---|
x+16= 32 | x+16= - 32 |
x=16 | x= - 48 |
The quadratic equation has two solutions, one negative and one positive. Because the dimensions of the pool cannot be negative, the negative solution is not an option for Vincenzo. Therefore, the dimensions of the pool are given by the positive solution. Width:&16 ft^2 Length:&16+32=48ft^2
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.
Substitute 1 into the equation for V and solve it by completing the square.
LHS+55^2=RHS+55^2
Split into factors
a^2-2ab+b^2=(a-b)^2
sqrt(LHS)=sqrt(RHS)
Calculate root
sqrt(a^2)=a
LHS+55=RHS+55
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.
The rocket has an initial vertical velocity of 32 feet per second. Additionally, the rocket will be launched from a height of 12 feet above the ground. The following quadratic equation describes the height of the rocket, where t is the time in seconds. h=-16t^2+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.No, see solution.
The square of a real number cannot be negative.
LHS-12=RHS-12
LHS * (-1)=RHS* (-1)
Distribute (-1)
(- a)(- b)=a* b
a(- b)=- a * b
a+(- b)=a-b
Split into factors
Factor out 16
.LHS /16.=.RHS /16.
Put minus sign in front of fraction
A * a/A= a
LHS+1=RHS+1
a^2-2ab+b^2=(a-b)^2
Add terms
In this lesson, quadratic expressions in the form x^2+bx were turned into perfect square trinomials in a process called completing the square. Similarly, quadratic equations were solved by completing the square. \begin{gathered} x^2+bx= c \\ \Downarrow \\ \underbrace{x^2+\dfrac{b}x {\color{#0000FF}{+\left(\dfrac{b}{2}\right)^2}}}_{\text{Perfect Square Trinomial}}= c{\color{#0000FF}{+\left(\dfrac{b}{2}\right)^2}} \end{gathered} 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.
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. x^2+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.
sqrt(LHS)=sqrt(RHS)
sqrt(a^2)=± a
Calculate root
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.
Determine the value of c that makes the expression a perfect square trinomial.
We are asked to find the value of c that makes the given expression a perfect square trinomial. To do this, we will first try to identify the value of b. In a quadratic expression, b is the coefficient of the linear term. x^2+ 36x+c To change the expression x^2+bx into a perfect square trinomial, calculate the square of half of the value of b. In this case, b= 36. Let's calculate it!
Following a similar reasoning as in Part A, let's identify the value of b.
s^2+ 14s+c
We have identified 14 to be the value of b. We will now calculate the square of half of 14.
In a similar fashion, let's find the value of the coefficient of the linear term for the given expression.
r^2+ 42r+c
Let's find the value of ( b2)^2.
Use the method of completing the square to solve the equation. When entering multiple roots, use the labels to change the number of entries.
We will find the solutions to the quadratic equation by completing the square. Observing the given equation, we can note that it is already written in the form x^2+bx=c. t^2+ 6t=55 To complete the square, we should add ( b2)^2 to both sides of the equation. In this case, b= 6. Let's first calculate ( b2)^2.
We will now add 9 to both sides of the equation. Then we will factor the resulting perfect square trinomial on the left-hand side.
We will take the square root of both sides of the equation. This will produce two linear equations whose solutions are also solutions of the quadratic equation.
Using the positive and negative signs will give us the solutions to the equation.
t=-3±8 | |
---|---|
t=-3+8 | t=-3-8 |
t=5 | t=-11 |
We have determined that t=5 and t=-11. These are the solutions to the quadratic equation.
We will begin by writing the given equation in the form x^2+bx=c. q^2-18q+32=0 ⇓ q^2-18q=-32 Note that b=-18. We will calculate ( b2)^2 that will be later added to both sides of the equation.
We will add 81 to both sides of the equation to form a perfect square trinomial on the left-hand side. This trinomial can later be factored as the square of a binomial.
Now, let's apply the square root to both sides of the equation to get two linear equations.
By taking the positive and negative signs, we will get the two solutions to the equation.
q=9±7 | |
---|---|
q=9+7 | q=9-7 |
q=16 | q=2 |
We have determined that q=16 and q=2 are the solutions of the quadratic equation.
Solve each equation by completing the square. Round to the nearest tenth if needed.
We will determine the solutions to the given equation by using the method of completing the square. To do so, we will first factor out 3.
Once the equation is in the form x^2+bx=c, we need to calculate the term that should be added to both sides of the equation to form a perfect square trinomial on its left-hand side. This is given by ( b2)^2. Note that in the resulting equation b= 12.
We will now add 36 to both sides of the equation in the form x^2+bx=c.
By applying the square root to both sides of the equation, we will obtain two linear equations whose solutions are also solutions of the quadratic equation.
We will separate the positive and negative cases and use a calculator to find the solutions to the equation.
x=-6±sqrt(27) | ||
---|---|---|
x=-6+sqrt(27) | x=-6-sqrt(27) | |
x≈-0.8 | x≈-11.2 |
Consider the given equation. 5x^2+10x-35=0 We will begin by factoring out 5 and then we will write the equation in the form x^2+bx=c.
Having the equation in the form x^2+bx=c, we will complete the square on the left-hand side. Let's first calculate the missing term ( b2)^2.
This term should be added now to both sides of the equation to produce a perfect square trinomial on the left hand side so that it can be factored as the square of a binomial.
We will now apply the square root to both sides of the equation to get two linear equations.
Finally, let's take the positive and negative cases and use a calculator to obtain the solutions to the equation.
x=-1±sqrt(8) | ||
---|---|---|
x ≈ -1+sqrt(8) | x ≈ -1-sqrt(8) | |
x≈1.8 | x≈-3.8 |
Use the method of completing the square to find the solutions of the equation.
The given equation is already in the form x^2+bx=c. x^2-5x=-6.25 This means we need to calculate the missing term, ( b2)^2, that should be added to both sides of the equation to obtain a perfect square trinomial on its left-hand side. In the given equation, b= -5.
We will add this term to both sides of the equation and factor the resulting perfect square trinomial.
We will apply the square root 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.
Consider the given equation.
x^2-4x=-6
We will calculate the missing term ( b2)^2. In this case, b= -4.
Let's add 4 to both sides of the equation to get a perfect square trinomial.
Next, we should apply the square root to both sides of the equation. However, we can 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.