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In perfect square trinomials with leading coefficient $1,$ there is a relationship between the coefficient of the $x-$term and the constant. ### Catch-Up and Review

$x_{2}+bx+(2b )_{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?
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.
*any* expression in the form $x_{2}+bx.$
*algebraic tiles*. Consider the following expression. *expand_more*
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$x_{2}+bx+(2b )_{2} $

This relationship can be used to form a perfect square trinomial by adding a constant $c$ to $x_{2}+bx⇒x_{2}+bx+c $

The process of finding the constant $c$ can be visualized by using $x_{2}+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.
This process is called completing the square. To complete the square for an expression algebraically, these steps can be followed.

1

Identify the Coefficient of the $x-$Term

For the given expression, the value of $b$ is $6.$

$x_{2}+6x $

2

Calculate $(2b )_{2}$

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

Add $(2b )_{2}$ to the expression to obtain a perfect square trinomial.

$x_{2}+bx⇒x_{2}+bx+(2b )_{2} $

In this case, $9$ should be added to $x_{2}+6x.$
$x_{2}+6x⇒x_{2}+6x+9 $

4

Factor the Perfect Square Trinomial

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 $x_{2}+bx.$

$x_{2}+bx+(2b )_{2}=(x+2b )_{2} $

This will be applied to the expression $x_{2}+6x+9.$
$x_{2}+6x+9$

SplitIntoFactors

Split into factors

$x_{2}+2⋅3x+9$

CommutativePropMult

Commutative Property of Multiplication

$x_{2}+2x⋅3+9$

FacPosPerfectSquare

$a_{2}+2ab+b_{2}=(a+b)_{2}$

$(x+3)_{2}$

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.

To solve a quadratic equation by completing the square, write the equation in the form $x_{2}+bx=c,$ 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.
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$2x_{2}+12x+4=0 $

To solve the equation by completing the square, these five steps can be followed.
1

Write the Equation in the Form $x_{2}+bx=c$

The Properties of Equality can be used jointly with inverse operations to rewrite the given equation in the form $x_{2}+bx=c.$
If the equation is already in the form $x_{2}+bx=c,$ this step is skipped.

$2x_{2}+12x+4=0$

SplitIntoFactors

Split into factors

$2x_{2}+2⋅6x+2⋅2=0$

FactorOut

Factor out $2$

$2(x_{2}+6x+2)=0$

DivEqn

$LHS/2=RHS/2$

$x_{2}+6x+2=0$

SubEqn

$LHS−2=RHS−2$

$x_{2}+6x=-2$

2

Complete the Square on the Left-Hand Side of the Equation

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+(2b )_{2}=c+(2b )_{2} $

In the equation found previously, $b$ is equal to $6.$ Therefore, $(26 )_{2}$ should be added to both sides.
$x_{2}+6x=-2⇓x_{2}+6x+(26 )_{2}=-2+(26 )_{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. 3

Factor the Perfect Square Trinomial

Next, factor the perfect square trinomial.

$x_{2}+6x+3_{2}=-2+3_{2}$

SplitIntoFactors

Split into factors

$x_{2}+2⋅x⋅3+3_{2}=-2+3_{2}$

FacPosPerfectSquare

$a_{2}+2ab+b_{2}=(a+b)_{2}$

$(x+3)_{2}=-2+3_{2}$

$(x+3)_{2}=7$

4

Take the Square Root of Both Sides of the Equation

The square root of both sides of the equation can now be taken to remove the exponent.

5

Solve for $x$

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 $x=-3−7 .$

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.

a $x_{2}+32x=768$

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.

a Use the formula for the area of a rectangle.

b The dimensions of the pool cannot be negative.

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=ℓwSubstitute 768=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 $x_{2}+bx=c.$

$x_{2}+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}⇔16_{2} $

This term needs to be added to both sides of the equation to produce an equivalent equation. $x_{2}+32x+16_{2}=768+16_{2} $

The perfect square trinomial can now be factored. $x_{2}+32x+16_{2}=768+16_{2}$

SplitIntoFactors

Split into factors

$x_{2}+2⋅x⋅16+16_{2}=768+16_{2}$

FacPosPerfectSquare

$a_{2}+2ab+b_{2}=(a+b)_{2}$

$(x+16)_{2}=768+16_{2}$

CalcPow

Calculate power

$(x+16)_{2}=768+256$

AddTerms

Add terms

$(x+16)_{2}=1024$

$(x+16)_{2}=1024$

SqrtEqn

$LHS =RHS $

$(x+16)_{2} =1024 $

$a_{2} =±a$

$x+16=±1024 $

CalcRoot

Calculate root

$x+16=±32$

$x+16=±32$ | |
---|---|

$x+16=32$ | $x+16=-32$ |

$x=16$ | $x=-48$ |

$Width:Length: 16ft_{2}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.

The following quadratic equation expresses the volume $V$ of liquid in the dispenser after $t$ minutes.$V=t_{2}−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"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":"minutes","answer":{"text":["55"]}}

Substitute $1$ into the equation for $V$ and solve it by completing the square.

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.
For this equation, the value of $b$ is $-110.$ With this information, $(2b )_{2}$ can be calculated.
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=t_{2}−110t+3026⇕t_{2}−110t+3026=1 $

The first step is to write the equation in the form $x+bx=c.$
$t_{2}−110t+3026=1$

SubEqn

$LHS−3026=RHS−3026$

$t_{2}−110t=1−3026$

SubTerm

Subtract term

$t_{2}−110t=-3025$

$(2-110 )_{2}⇔(-55)_{2}=55_{2} $

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.
$t_{2}−110t=-3025$

AddEqn

$LHS+55_{2}=RHS+55_{2}$

$t_{2}−110t+55_{2}=-3025+55_{2}$

SplitIntoFactors

Split into factors

$t_{2}−2⋅t⋅55+55_{2}=-3025+55_{2}$

FacNegPerfectSquare

$a_{2}−2ab+b_{2}=(a−b)_{2}$

$(t−55)_{2}=-3025+55_{2}$

$(t−55)_{2}=0$

$(t−55)_{2}=0$

Solve for $t$

SqrtEqn

$LHS =RHS $

$(t−55)_{2} =0 $

CalcRoot

Calculate root

$(t−55)_{2} =0$

SqrtPowToNumber

$a_{2} =a$

$t−55=0$

AddEqn

$LHS+55=RHS+55$

$t=55$

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.

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=-16t_{2}+32t+12⇕-16t_{2}+32t+12=60 $

The equation must first be written in the form $x_{2}+bx=c$ to solve it by completing the square. $-16t_{2}+32t+12=60$

SubEqn

$LHS−12=RHS−12$

$-16t_{2}+32t=48$

Rewrite

MultEqn

$LHS⋅(-1)=RHS⋅(-1)$

$(-16t_{2}+32t)(-1)=48(-1)$

Distr

Distribute $(-1)$

$(-16t_{2})(-1)+(32t)(-1)=48(-1)$

MultNegNegTwoPar

$(-a)(-b)=a⋅b$

$16t_{2}+(32t)(-1)=48(-1)$

MultPosNeg

$a(-b)=-a⋅b$

$16t_{2}+(-32t)=-48$

AddNeg

$a+(-b)=a−b$

$16t_{2}−32t=-48$

SplitIntoFactors

Split into factors

$16⋅t_{2}−16⋅2⋅t=-16⋅3$

FactorOut

Factor out $16$

$16(t_{2}−2t)=-16⋅3$

DivEqn

$LHS/16=RHS/16$

$t_{2}−2t=16-16⋅3 $

MoveNegNumToFrac

Put minus sign in front of fraction

$t_{2}−2t=-1616⋅3 $

DenomMultFracToNumber

$A⋅Aa =a$

$t_{2}−2t=-3$

$(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.
$t_{2}−2t=-3$

AddEqn

$LHS+1=RHS+1$

$t_{2}−2t+1=-3+1$

FacNegPerfectSquare

$a_{2}−2ab+b_{2}=(a−b)_{2}$

$(t−1)_{2}=-3+1$

AddTerms

Add terms

$(t−1)_{2}=-2$

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.
### Hint

### Solution

$x_{2}+bx=c⇓Perfect Square Trinomialx_{2}+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.
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. a What is the correct number for the second to last digit?

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["4"]}}

b What is the last digit of the combination?

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["6"]}}

a Consider the the lock contains only non-negative numbers.

b Add $2$ to the second to last number of the code.