Unit 8 Plan - Algebra 1

Title	Assessment
Lesson 1 Finding Unknown Inputs	Interpreting a Solution A framed picture has a total area $y$ , in square inches. The thickness of the frame is represented by $x$ , in inches. The equation $y=(8+2x)(10+2x)$ relates these two variables. What are the length and width of the picture without the frame? What would a solution to the equation $100=(8+2x)(10+2x)$ mean in this situation? Show Solution 8 inches and 10 inches A solution would represent the thickness of a frame that results in a total area of 100 square inches.
Lesson 2 When and Why Do We Write Quadratic Equations?	The Movie Theater A movie theater models the revenue from ticket sales in one day as a function of the ticket price, $p$ . Here are two expressions defining the same revenue function. $\displaystyle p(120-4p)$ $\displaystyle 120p-4p^2$ According to this model, how high would the ticket price have to be for the theater to make $0 in revenue? Explain your reasoning. What equation can you write to find out what ticket price(s) would allow the theater to make $600 in revenue? Show Solution $30. Sample reasoning: If $p(120-4p)=0$ , then either $p=0$ or $120-4p=0$ . If the latter is a true equation, $p$ must be 30. $p(120-4p)=600$ or $120p - 4p^2=600$
Section A Check Section A Checkpoint	Problem 1 An artist is creating a piece that will begin as a square piece of paper that is 17 inches on each side. They will print text on a portion of the paper, leaving some white space on each side. The spaces on the top and bottom of the paper will be equal and the spaces to the left and right will each be twice as much as on the top and bottom. The area of the printed text is 130 square inches. Write an equation that represents the printed area. The approximate solutions are $x = 1.82$ and $x = 10.93$ . Do these solutions make sense in this situation? Explain your reasoning. Show Solution $(17-2x)(17-4x)=130$ 1.82 makes sense, but 10.93 does not. Sample reasoning: When $x = 1.82$ , one side of the printed area will be 13.36 inches ( $17 - 2\boldcdot 1.82 = 13.36$ ) and the other side of the printed area will be 9.72 inches ( $17-4\boldcdot 1.82 = 9.72$ ). If $x = 10.93$ , then there is more than 20 inches of margin on the top and bottom together, but the page is only 17 inches tall.

Lesson 3 Solving Quadratic Equations by Reasoning	Find Both Solutions Find both solutions to the equation $100+(n-2)^2=149$ . Explain or show your reasoning. Show Solution 9 and -5. Sample reasoning: 100 plus a squared number is 149. That squared number must be 49 and the number must be 7 or -7. If $n-2=7$ , then $n$ is 9. If $n-2=\text-7$ , then $n= \text-5$ .
Lesson 4 Solving Quadratic Equations with the Zero Product Property	Solve This Equation! Find all solutions to $(x+5)(2x-3)=0$ . Explain or show your reasoning. Show Solution -5 and $\frac32$ . Sample reasoning: By the zero product property, either $x+5=0$ or $2x-3=0$ , so $x=\text-5$ or $x=\frac32$ .
Lesson 5 How Many Solutions?	Two, One, or None? For each quadratic equation, decide whether it has two solutions, one solution, or no solutions. Explain how you know. $x^2=\text-16$ $x(x+2)=0$ $(x-3)(x-3)=0$ Show Solution No solutions. Sample reasoning: There are no numbers we can square to get a negative product. The graph of $y=x^2+16$ has no $x$ -intercepts. Two solutions. Sample reasoning: Both 0 and -2 make the equation true. The graph of $y=x(x+2)$ has two $x$ -intercepts. One solution. Sample reasoning: Only 3 makes the equation true. The graph of $y=(x-3)(x-3)$ has one $x$ -intercept.
Lesson 6 Rewriting Quadratic Expressions in Factored Form (Part 1)	The Unknown Numbers Here are pairs of equivalent expressions—one in standard form and the other in factored form. Find the numbers that go in the boxes. $x^2 + \boxed{\phantom{300}}x +\boxed{\phantom{300}}$ and $(x+1)(x+17)$ $x^2 + 9x + 14$ and $(x+2)(x+\boxed{\phantom{300}})$ $x^2 - 7x + 10$ and $(x-2)(x-\boxed{\phantom{300}})$ $x^2 - 9x + 20$ and $(x-\boxed{\phantom{300}})(x-\boxed{\phantom{300}})$ Show Solution 18 and 17. The completed expression is $x^2+18x+17$ . 7. The completed expression is $(x+2)(x+7)$ . 5. The completed expression is $(x-2)(x-5)$ . 4 and 5. The completed expression is $(x-4)(x-5)$ .
Lesson 7 Rewriting Quadratic Expressions in Factored Form (Part 2)	The Unknown Symbols Here are pairs of equivalent expressions in standard form and factored form. Find the unknown symbols and numbers. $x^2 \boxed{\phantom{30}}\>16x \> \boxed{\phantom{30}}\> 17$ and $(x+1)(x-17)$ $x^2 \boxed{\phantom{30}} \> 16x \> \boxed{\phantom{30}}\> 17$ and $(x-1)(x+17)$ $x^2 + 3x - 28$ and $(x + \boxed{\phantom{3000}})(x - \boxed{\phantom{3000}})$ $x^2 - 12x - 28$ and $(x \> \boxed{\phantom{30}} \> \boxed{\phantom{3000}})(x \> \boxed{\phantom{30}} \>\boxed{\phantom{3000}})$ Show Solution $-$ and $-$ . The completed expression is $x^2 - 16x - 17$ . $+$ and $-$ . The completed expression is $x^2 +16x - 17$ . 7 and 4. The completed expression is $(x + 7)(x - 4)$ . $-14$ and $+2$ . The completed expression is $(x - 14)(x + 2)$ .
Lesson 8 Rewriting Quadratic Expressions in Factored Form (Part 3)	Can These Be Rewritten in Factored Form? Write each expression in factored form. If it is not possible, write “not possible.” $a^2-36$ $49-25b^2$ $c^2+9$ $\frac{100}{81}-16d^2$ Show Solution $(a+6)(a-6)$ $(7+5b)(7-5b)$ not possible $(\frac{10}{9}+4d)(\frac{10}{9}-4d)$
Lesson 9 Solving Quadratic Equations by Using Factored Form	Conquering More Equations Solve each equation by rewriting it in factored form and using the zero product property. Show your reasoning. $x^2+12x+11=0$ $x^2-3=1$ $x^2-6x+7=\text-2$ Show Solution -1 and -11. The equation $x^2 + 12x + 11=0$ can be written as $(x+11)(x+1)=0$ , which means $x+11=0$ or $x+1=0$ . -2 and 2. The equation $x^2 - 3 = 1$ can be written as $x^2-4=0$ , which is $(x+2)(x-2)=0$ , which means $x+2=0$ or $x-2=0$ . 3. The equation $x^2 - 6x + 7 = \text-2$ can be written as $x^2 - 6x + 9 = 0$ , which is $(x-3)(x-3)=0$ , which means $x-3=0$ .
Section B Check Section B Checkpoint	Problem 1 Solve each quadratic equation. $(x+3)(5-x) = 0$ $(2x+1)(x-6) = 0$ $6 = 6 + (x-1)(x+4)$ Show Solution $x = \text{-}3$ and $x= 5$ $x = \text{-}\frac{1}{2}$ and $x = 6$ $x = 1$ and $x = \text{-}4$ Problem 2 Rewrite each expression in factored form. $x^2 + 4x$ $x^2 + 3x - 4$ $x^2 - 81$ $2x^2 +7x+3$ Show Solution $x(x+4)$ $(x-1)(x+4)$ $(x-9)(x+9)$ $(2x-1)(x+3)$

Lesson 11 What Are Perfect Squares?	A Perfect Square Explain why it makes sense to call the expression $x^2+20x+100$ a “perfect square.” Solve $x^2+20x+100=81$ . Show Solution Sample response: It is equivalent to $(x+10)^2$ , so it is a factor squared. -1 and -19
Lesson 12 Completing the Square (Part 1)	Make It a Perfect Square What could be added to each expression to make it a perfect square? $x^2 + 12x$ $x^2 - 6x + 1$ $x^2 + 14x - 10$ Solve the equation $x^2 -16x = \text-60$ by completing the square. Show your reasoning. Show Solution 36 8 59 6 and 10. Sample reasoning: $\displaystyle \begin {align} x^2-16x &= \text-60 \\ x^2-16x+64 &=\text-60 + 64\\x^2-16x+64 &=4 \\ (x-8)^2 &=4\\ x-8=2 \quad & \text{or} \quad x-8=\text-2\\ x=10 \quad & \text{or} \quad x=6 \end{align}$
Lesson 13 Completing the Square (Part 2)	How Did We Get Those Solutions? The solutions to this equation are $\frac34$ and $\frac14$ . Show how to find those solutions by completing the square. $\displaystyle x^2 - x + \frac{3}{16}=0$ Show Solution $\frac34$ and $\frac14$ . Sample reasoning: $\displaystyle \begin {align} x^2 - x + \frac{3}{16}&=0\\ x^2 - x + \frac14 &= \frac{1}{16}\\ (x - \frac12)^2 &= \frac{1}{16}\\ \\ x - \frac12 = \frac14 \quad &\text{or} \quad x - \frac12 =\text- \frac14\\x = \frac34 \quad &\text{or} \quad x = \frac14 \end {align}$
Lesson 15 Quadratic Equations with Irrational Solutions	Finding Exact Solutions For the equation $x^2+14x=\text-46$ , the approximate solutions are -8.732 and -5.268. Find the exact solutions by completing the square. Show your reasoning. Show Solution $\text-7 \pm \sqrt3$ Sample reasoning: $\displaystyle \begin {align} x^2+14x &= \text-46 \\ x^2+14x+49 &=\text-46 + 49\\x^2+14x+49 &=3 \\ (x+7)^2 &=3\\ x+7&=\pm\sqrt3 \\ x&=\text-7\pm\sqrt3 \end{align}$
Section C Check Section C Checkpoint	Problem 1 Complete the square to solve the equations. Give exact solutions. $x^2 + 2x = 3$ $x^2 + 6x + 3 =0$ Show Solution $x = 1$ and $x = \text{-}3$ . Sample reasoning: Completing the square leads to $(x+1)^2 = 4$ . This means that $x + 1 = 2$ and $x + 1 = \text{-}2$ . $x = \text{-}3\pm\sqrt{6}$ . Sample reasoning: Completing the square leads to $(x+3)^2 = 6$ . This means that $x + 3 = \pm \sqrt{6}$ . Problem 2 Select all of the equations that are equivalent to $x = 3 \pm \sqrt{2}$ . A. $x = \sqrt{5}$ B. $x = \sqrt{\text{-}1}$ C. $x = 3 + \sqrt{2}$ D. $x = \sqrt{\pm 6}$ E. $x = 3 - \sqrt{2}$ Show Solution C, E

Lesson 17 Applying the Quadratic Formula (Part 1)	Tennis Ball Up, Tennis Ball Down Function $h$ gives the height of a tennis ball, in feet, $t$ seconds after it is tossed straight up in the air. The equation $h(t)= \text-16t^2+12t+10$ defines function $h$ . Write and solve an equation to find when the ball hits the ground. Show your reasoning. Show Solution The equation $\text-16t^2+12t+10=0$ can be rewritten as $(4t+2)(\text-4t + 5)=0$ . By the zero product property, $4t+2=0$ or $(\text-4t + 5)=0$ , so $t=\text-\frac12$ or $t= \frac54$ . Only the positive solution makes sense here, so the ball hits the ground 1.25 seconds after being tossed up.
Lesson 18 Applying the Quadratic Formula (Part 2)	Where Did It Go Wrong? Here is someone’s solution to the equation $3x^2 - 4x = 20$ . $\begin {align}a &=3 \quad b =\text-4 \quad c = 20\\ \\x &=\dfrac{\text-b \pm \sqrt{b^2-4ac}}{2a}\\x &=\dfrac{\text-4 \pm \sqrt{(\text-4)^2-4(3)(20)}}{2(3)}\\ t &=\dfrac{\text-4 \pm \sqrt{\text-16-240}}{6} \\x &=\dfrac{\text-4 \pm \sqrt{\text-256}}{6}\\ &\text{No solutions}\end{align}$ Identify as many errors as you can and briefly explain each error. Show Solution Sample response: $b$ is -4, so the value $\text-b$ should be $\text-(\text-4)$ , which is 4. When the equation is rewritten as $ax^2+bx+c=0$ , the constant term $c$ is -20, not 20, so the last part of the numerator should be $\sqrt{(\text-4)^2-4(3)(\text-20)}$ . $(\text-4)^2$ should be 16. Squaring a negative number gives a positive number. If the earlier issues were fixed, then the values under the square root symbol would be $16+240$ , or 256.
Lesson 20 Rational and Irrational Solutions	What Kind of Solutions? Decide whether the solutions to each equation are rational or irrational. Explain your reasoning. $(x+5)^2=9$ $(x+5)^2=10$ Someone says that if you add two irrational numbers, you will always get an irrational sum. How can you convince the person that they are wrong? Show Solution Explanations vary. A likely approach is to find the solutions of $\text-5 \pm \sqrt9$ and $\text-5 \pm \sqrt{10}$ and note that 9 is a perfect square, so its square roots are $\pm3$ , but 10 is not a perfect square. rational irrational Sample response: We can show a counterexample—for example, $\sqrt3 + \text-\sqrt3 = 0$ .
Section D Check Section D Checkpoint	Problem 1 Use the quadratic formula to find exact solutions to these equations. $x = \dfrac{\text-b \pm \sqrt{b^2-4ac}} { 2a}$ $3x^2 - 2x - 1 = 0$ $x^2 + 4x = 1$ Show Solution $x = \text{-}\frac{1}{3}$ and $x = 1$ $x = \text{-}2 \pm \sqrt{5}$ Problem 2 Classify each value as rational or irrational. Explain your reasoning. $2 \sqrt{9}$ $3 - \sqrt{10}$ $\sqrt{1} \boldcdot \sqrt{2}$ Show Solution Rational. Sample reasoning: $2 \sqrt{9} = 2\boldcdot 3 = 6$ Irrational. Sample reasoning: The sum of a rational number and an irrational number is irrational. Irrational. Sample reasoning: The product of a nonzero rational number and an irrational number is irrational.

Section E Check Section E Checkpoint	Problem 1 For each quadratic function, if it is not in vertex form, rewrite the equation in vertex form. Then, find the vertex, and state whether it is a maximum or minimum. $f(x) = x^2 + 10x -2$ $g(x) = \text{-}(x+2)^2+5$ Show Solution $f(x) = (x+5)^2 -27$ . Vertex: $(\text{-}5,\text{-}27)$ , minimum It is in vertex form. Vertex: $(\text{-}2,5)$ , maximum

Lesson 24 Using Quadratic Equations to Model Situations and Solve Problems	Profit from a River Cruise A travel company uses a quadratic function to model the profit, in dollars, that it expects to earn from selling tickets for a river cruise at $d$ dollars per person. The expression $\text-d^2 +100d - 900$ defines this function. Without graphing, find the price that would generate the maximum profit. Then, determine that maximum profit. Show Solution The price of $50 per ticket would generate a maximum profit of $1,600. Sample reasoning: $\text-d^2 +100d - 900$ can be rewritten in vertex form as $\text-(d-50)^2 + 1,600$ , so the vertex of the graph is at $(50, 1,600)$ . This means the maximum profit, $1,600, can be expected when tickets are priced at $50 each.

Lesson 1

Finding Unknown Inputs

Interpreting a Solution

A framed picture has a total area $y$ , in square inches. The thickness of the frame is represented by $x$ , in inches. The equation $y=(8+2x)(10+2x)$ relates these two variables.

What are the length and width of the picture without the frame?
What would a solution to the equation $100=(8+2x)(10+2x)$ mean in this situation?

Show Solution

8 inches and 10 inches
A solution would represent the thickness of a frame that results in a total area of 100 square inches.

Lesson 2

When and Why Do We Write Quadratic Equations?

The Movie Theater

A movie theater models the revenue from ticket sales in one day as a function of the ticket price, $p$ . Here are two expressions defining the same revenue function.

$\displaystyle p(120-4p)$

$\displaystyle 120p-4p^2$

According to this model, how high would the ticket price have to be for the theater to make $0 in revenue? Explain your reasoning.
What equation can you write to find out what ticket price(s) would allow the theater to make $600 in revenue?

Show Solution

$30. Sample reasoning: If $p(120-4p)=0$ , then either $p=0$ or $120-4p=0$ . If the latter is a true equation, $p$ must be 30.
$p(120-4p)=600$ or $120p - 4p^2=600$

Section A Check

Section A Checkpoint

Problem 1

An artist is creating a piece that will begin as a square piece of paper that is 17 inches on each side. They will print text on a portion of the paper, leaving some white space on each side. The spaces on the top and bottom of the paper will be equal and the spaces to the left and right will each be twice as much as on the top and bottom.

The area of the printed text is 130 square inches.

Write an equation that represents the printed area.
The approximate solutions are $x = 1.82$ and $x = 10.93$ . Do these solutions make sense in this situation? Explain your reasoning.

Show Solution

$(17-2x)(17-4x)=130$
1.82 makes sense, but 10.93 does not. Sample reasoning: When $x = 1.82$ , one side of the printed area will be 13.36 inches ( $17 - 2\boldcdot 1.82 = 13.36$ ) and the other side of the printed area will be 9.72 inches ( $17-4\boldcdot 1.82 = 9.72$ ). If $x = 10.93$ , then there is more than 20 inches of margin on the top and bottom together, but the page is only 17 inches tall.

Lesson 3

Solving Quadratic Equations by Reasoning

Find Both Solutions

Find both solutions to the equation $100+(n-2)^2=149$ . Explain or show your reasoning.

Show Solution

9 and -5. Sample reasoning: 100 plus a squared number is 149. That squared number must be 49 and the number must be 7 or -7. If $n-2=7$ , then $n$ is 9. If $n-2=\text-7$ , then $n= \text-5$ .

Lesson 4

Solving Quadratic Equations with the Zero Product Property

Solve This Equation!

Find all solutions to $(x+5)(2x-3)=0$ . Explain or show your reasoning.

Show Solution

-5 and $\frac32$ . Sample reasoning: By the zero product property, either $x+5=0$ or $2x-3=0$ , so $x=\text-5$ or $x=\frac32$ .

Lesson 5

How Many Solutions?

Two, One, or None?

For each quadratic equation, decide whether it has two solutions, one solution, or no solutions. Explain how you know.

$x^2=\text-16$
$x(x+2)=0$
$(x-3)(x-3)=0$

Show Solution

No solutions. Sample reasoning:
- There are no numbers we can square to get a negative product.
- The graph of $y=x^2+16$ has no $x$ -intercepts.
Two solutions. Sample reasoning:
- Both 0 and -2 make the equation true.
- The graph of $y=x(x+2)$ has two $x$ -intercepts.
One solution. Sample reasoning:
- Only 3 makes the equation true.
- The graph of $y=(x-3)(x-3)$ has one $x$ -intercept.

Lesson 6

Rewriting Quadratic Expressions in Factored Form (Part 1)

The Unknown Numbers

Here are pairs of equivalent expressions—one in standard form and the other in factored form. Find the numbers that go in the boxes.

$x^2 + \boxed{\phantom{300}}x +\boxed{\phantom{300}}$ and $(x+1)(x+17)$
$x^2 + 9x + 14$ and $(x+2)(x+\boxed{\phantom{300}})$
$x^2 - 7x + 10$ and $(x-2)(x-\boxed{\phantom{300}})$
$x^2 - 9x + 20$ and $(x-\boxed{\phantom{300}})(x-\boxed{\phantom{300}})$

Show Solution

18 and 17. The completed expression is $x^2+18x+17$ .
7. The completed expression is $(x+2)(x+7)$ .
5. The completed expression is $(x-2)(x-5)$ .
4 and 5. The completed expression is $(x-4)(x-5)$ .

Lesson 7

Rewriting Quadratic Expressions in Factored Form (Part 2)

The Unknown Symbols

Here are pairs of equivalent expressions in standard form and factored form. Find the unknown symbols and numbers.

$x^2 \boxed{\phantom{30}}\>16x \> \boxed{\phantom{30}}\> 17$ and $(x+1)(x-17)$
$x^2 \boxed{\phantom{30}} \> 16x \> \boxed{\phantom{30}}\> 17$ and $(x-1)(x+17)$
$x^2 + 3x - 28$ and $(x + \boxed{\phantom{3000}})(x - \boxed{\phantom{3000}})$
$x^2 - 12x - 28$ and $(x \> \boxed{\phantom{30}} \> \boxed{\phantom{3000}})(x \> \boxed{\phantom{30}} \>\boxed{\phantom{3000}})$

Show Solution

$-$ and $-$ . The completed expression is $x^2 - 16x - 17$ .
$+$ and $-$ . The completed expression is $x^2 +16x - 17$ .
7 and 4. The completed expression is $(x + 7)(x - 4)$ .
$-14$ and $+2$ . The completed expression is $(x - 14)(x + 2)$ .

Lesson 8

Rewriting Quadratic Expressions in Factored Form (Part 3)

Can These Be Rewritten in Factored Form?

Write each expression in factored form. If it is not possible, write “not possible.”

$a^2-36$
$49-25b^2$
$c^2+9$
$\frac{100}{81}-16d^2$

Show Solution

$(a+6)(a-6)$
$(7+5b)(7-5b)$
not possible
$(\frac{10}{9}+4d)(\frac{10}{9}-4d)$

Lesson 9

Solving Quadratic Equations by Using Factored Form

Conquering More Equations

Solve each equation by rewriting it in factored form and using the zero product property. Show your reasoning.

$x^2+12x+11=0$
$x^2-3=1$
$x^2-6x+7=\text-2$

Show Solution

-1 and -11. The equation $x^2 + 12x + 11=0$ can be written as $(x+11)(x+1)=0$ , which means $x+11=0$ or $x+1=0$ .
-2 and 2. The equation $x^2 - 3 = 1$ can be written as $x^2-4=0$ , which is $(x+2)(x-2)=0$ , which means $x+2=0$ or $x-2=0$ .
3. The equation $x^2 - 6x + 7 = \text-2$ can be written as $x^2 - 6x + 9 = 0$ , which is $(x-3)(x-3)=0$ , which means $x-3=0$ .

Section B Check

Section B Checkpoint

Problem 1

Solve each quadratic equation.

$(x+3)(5-x) = 0$
$(2x+1)(x-6) = 0$
$6 = 6 + (x-1)(x+4)$

Show Solution

$x = \text{-}3$ and $x= 5$
$x = \text{-}\frac{1}{2}$ and $x = 6$
$x = 1$ and $x = \text{-}4$

Problem 2

Rewrite each expression in factored form.

$x^2 + 4x$
$x^2 + 3x - 4$
$x^2 - 81$
$2x^2 +7x+3$

Show Solution

$x(x+4)$
$(x-1)(x+4)$
$(x-9)(x+9)$
$(2x-1)(x+3)$

Lesson 11

What Are Perfect Squares?

A Perfect Square

Explain why it makes sense to call the expression $x^2+20x+100$ a “perfect square.”
Solve $x^2+20x+100=81$ .

Show Solution

Sample response: It is equivalent to $(x+10)^2$ , so it is a factor squared.
-1 and -19

Lesson 12

Completing the Square (Part 1)

Make It a Perfect Square

What could be added to each expression to make it a perfect square?
1. $x^2 + 12x$
2. $x^2 - 6x + 1$
3. $x^2 + 14x - 10$
Solve the equation $x^2 -16x = \text-60$ by completing the square. Show your reasoning.

Show Solution

1. 36
2. 8
3. 59
6 and 10. Sample reasoning:
$\displaystyle \begin {align} x^2-16x &= \text-60 \\ x^2-16x+64 &=\text-60 + 64\\x^2-16x+64 &=4 \\ (x-8)^2 &=4\\ x-8=2 \quad & \text{or} \quad x-8=\text-2\\ x=10 \quad & \text{or} \quad x=6 \end{align}$

Lesson 13

Completing the Square (Part 2)

How Did We Get Those Solutions?

The solutions to this equation are $\frac34$ and $\frac14$ . Show how to find those solutions by completing the square.

$\displaystyle x^2 - x + \frac{3}{16}=0$

Show Solution

$\frac34$ and $\frac14$ . Sample reasoning:

$\displaystyle \begin {align} x^2 - x + \frac{3}{16}&=0\\ x^2 - x + \frac14 &= \frac{1}{16}\\ (x - \frac12)^2 &= \frac{1}{16}\\ \\ x - \frac12 = \frac14 \quad &\text{or} \quad x - \frac12 =\text- \frac14\\x = \frac34 \quad &\text{or} \quad x = \frac14 \end {align}$

Lesson 15

Quadratic Equations with Irrational Solutions

Finding Exact Solutions

For the equation $x^2+14x=\text-46$ , the approximate solutions are -8.732 and -5.268.

Find the exact solutions by completing the square. Show your reasoning.

Show Solution

$\text-7 \pm \sqrt3$

Sample reasoning:

$\displaystyle \begin {align} x^2+14x &= \text-46 \\ x^2+14x+49 &=\text-46 + 49\\x^2+14x+49 &=3 \\ (x+7)^2 &=3\\ x+7&=\pm\sqrt3 \\ x&=\text-7\pm\sqrt3 \end{align}$

Section C Check

Section C Checkpoint

Problem 1

Complete the square to solve the equations. Give exact solutions.

$x^2 + 2x = 3$
$x^2 + 6x + 3 =0$

Show Solution

$x = 1$ and $x = \text{-}3$ . Sample reasoning: Completing the square leads to $(x+1)^2 = 4$ . This means that $x + 1 = 2$ and $x + 1 = \text{-}2$ .
$x = \text{-}3\pm\sqrt{6}$ . Sample reasoning: Completing the square leads to $(x+3)^2 = 6$ . This means that $x + 3 = \pm \sqrt{6}$ .

Problem 2

Select all of the equations that are equivalent to

x = 3 \pm \sqrt{2}

.

A.

x = \sqrt{5}

B.

x = \sqrt{\text{-}1}

C.

x = 3 + \sqrt{2}

D.

x = \sqrt{\pm 6}

E.

x = 3 - \sqrt{2}

Show Solution

C, E

Lesson 17

Applying the Quadratic Formula (Part 1)

Tennis Ball Up, Tennis Ball Down

Function $h$ gives the height of a tennis ball, in feet, $t$ seconds after it is tossed straight up in the air. The equation $h(t)= \text-16t^2+12t+10$ defines function $h$ .

Write and solve an equation to find when the ball hits the ground. Show your reasoning.

Show Solution

The equation $\text-16t^2+12t+10=0$ can be rewritten as $(4t+2)(\text-4t + 5)=0$ . By the zero product property, $4t+2=0$ or $(\text-4t + 5)=0$ , so $t=\text-\frac12$ or $t= \frac54$ . Only the positive solution makes sense here, so the ball hits the ground 1.25 seconds after being tossed up.

Lesson 18

Applying the Quadratic Formula (Part 2)

Where Did It Go Wrong?

Here is someone’s solution to the equation $3x^2 - 4x = 20$ .

$\begin {align}a &=3 \quad b =\text-4 \quad c = 20\\ \\x &=\dfrac{\text-b \pm \sqrt{b^2-4ac}}{2a}\\x &=\dfrac{\text-4 \pm \sqrt{(\text-4)^2-4(3)(20)}}{2(3)}\\ t &=\dfrac{\text-4 \pm \sqrt{\text-16-240}}{6} \\x &=\dfrac{\text-4 \pm \sqrt{\text-256}}{6}\\ &\text{No solutions}\end{align}$

Identify as many errors as you can and briefly explain each error.

Show Solution

Sample response:

$b$ is -4, so the value $\text-b$ should be $\text-(\text-4)$ , which is 4.
When the equation is rewritten as $ax^2+bx+c=0$ , the constant term $c$ is -20, not 20, so the last part of the numerator should be $\sqrt{(\text-4)^2-4(3)(\text-20)}$ .
$(\text-4)^2$ should be 16. Squaring a negative number gives a positive number.
If the earlier issues were fixed, then the values under the square root symbol would be $16+240$ , or 256.

Lesson 20

Rational and Irrational Solutions

What Kind of Solutions?

Decide whether the solutions to each equation are rational or irrational. Explain your reasoning.
1. $(x+5)^2=9$
2. $(x+5)^2=10$
Someone says that if you add two irrational numbers, you will always get an irrational sum. How can you convince the person that they are wrong?

Show Solution

Explanations vary. A likely approach is to find the solutions of $\text-5 \pm \sqrt9$ and $\text-5 \pm \sqrt{10}$ and note that 9 is a perfect square, so its square roots are $\pm3$ , but 10 is not a perfect square.
1. rational
2. irrational
Sample response: We can show a counterexample—for example, $\sqrt3 + \text-\sqrt3 = 0$ .

Section D Check

Section D Checkpoint

Problem 1

Use the quadratic formula to find exact solutions to these equations.

$x = \dfrac{\text-b \pm \sqrt{b^2-4ac}} { 2a}$

$3x^2 - 2x - 1 = 0$
$x^2 + 4x = 1$

Show Solution

$x = \text{-}\frac{1}{3}$ and $x = 1$
$x = \text{-}2 \pm \sqrt{5}$

Problem 2

Classify each value as rational or irrational. Explain your reasoning.

$2 \sqrt{9}$
$3 - \sqrt{10}$
$\sqrt{1} \boldcdot \sqrt{2}$

Show Solution

Rational. Sample reasoning: $2 \sqrt{9} = 2\boldcdot 3 = 6$
Irrational. Sample reasoning: The sum of a rational number and an irrational number is irrational.
Irrational. Sample reasoning: The product of a nonzero rational number and an irrational number is irrational.

Section E Check

Section E Checkpoint

Problem 1

For each quadratic function, if it is not in vertex form, rewrite the equation in vertex form. Then, find the vertex, and state whether it is a maximum or minimum.

$f(x) = x^2 + 10x -2$
$g(x) = \text{-}(x+2)^2+5$

Show Solution

$f(x) = (x+5)^2 -27$ . Vertex: $(\text{-}5,\text{-}27)$ , minimum
It is in vertex form. Vertex: $(\text{-}2,5)$ , maximum

Lesson 24

Using Quadratic Equations to Model Situations and Solve Problems

Profit from a River Cruise

A travel company uses a quadratic function to model the profit, in dollars, that it expects to earn from selling tickets for a river cruise at $d$ dollars per person. The expression $\text-d^2 +100d - 900$ defines this function.

Without graphing, find the price that would generate the maximum profit. Then, determine that maximum profit.

Show Solution

The price of $50 per ticket would generate a maximum profit of $1,600. Sample reasoning: $\text-d^2 +100d - 900$ can be rewritten in vertex form as $\text-(d-50)^2 + 1,600$ , so the vertex of the graph is at $(50, 1,600)$ . This means the maximum profit, $1,600, can be expected when tickets are priced at $50 each.

Unit 8 Quadratic Equations — Unit Plan