Lesson 17
Applying the Quadratic Formula (Part 1)
17.1: No Solutions for You! (5 minutes)
Warmup
Previously, students have seen that some quadratic functions have no zeros and that some quadratic equations have no solutions. In this warmup, they recall that when a solution is a square root of a negative number, the equation has no solutions. Note that students don’t yet know about any numbers that aren’t real at this point, so it is unnecessary to specify “no real solutions.” (To students, the word “real” would seem like an extra word added for no reason.)
Launch
Arrange students in groups of 2. Give students quiet think time, and then time to share their thinking with a partner.
Student Facing
Here is an example of someone solving a quadratic equation that has no solutions:
\(\displaystyle \begin {align} (x+3)^2+9 &=0\\ (x+3)^2 &=\text9\\ x+3 &=\pm \sqrt{\text9} \end {align}\)
 Study the example. At what point did you realize the equation had no solutions?
 Explain how you know the equation \(49+x^2=0\) has no solutions.
Student Response
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Activity Synthesis
Select students to share their responses and reasoning.
If not mentioned in students’ explanations, point out that we can look at and reason about the structure of these equations to tell which ones have no solutions, without taking any steps to solve. For example, the equation \((x+3)^2 + 9=0\) means “a square plus 9 equals 0.” Here are two ways to reason about its structure:
 For two numbers to add up to 0, they have to be opposites. Since the number 9 is positive, the other number must be negative, which a square cannot be.
 A square is always positive, so a square plus a positive number must also be positive and can never equal zero.
These lines of reasoning also allow us to see that \(49+x^2=0\) has no solutions.
17.2: The Potato and the Pumpkin (15 minutes)
Activity
By now, students are pretty familiar with quadratic functions that model the height of objects as a function of time after being launched up or being dropped. They have found the zeros of such functions graphically, by identifying horizontal intercepts, as well as algebraically, by writing equations of the form \(\text{expression in factored form}=0\) and applying the zero product property.
Students have also used graphs to estimate input values that yield nonzero output values. Prior to learning about completing the square or the quadratic formula, however, they did not have the tools to solve for such inputs algebraically. In this activity, students apply the knowledge and skills they recently developed to solve contextual problems that they couldn’t previously solve without graphing.
Because algebraic reasoning is the aim of this activity, graphing technology should not be used to solve the equations. It could be used to verify solutions during the activity synthesis, however.
If time is limited, assign the first question to half of the class and the second question to the other half. Alternatively, ask students to choose one question to answer.
Launch
Remind students that in earlier lessons they encountered two functions that modeled the height of a launched object as a function of time. They solved problems about the functions by graphing. Tell students that they now have additional strategies at their disposal and ask them to solve these problems without graphing. Provide access to calculators. Tell students to use them only for numerical computations. If time is limited, consider asking students to answer only the first set of questions.
Supports accessibility for: Conceptual processing; Memory
Student Facing
Answer each question without graphing. Explain or show your reasoning.

The equation \(h(t) = \text16t^2 + 80t + 64\) represented the height, in feet, of a potato \(t\) seconds after it has been launched.
 Write an equation that can be solved to find when the potato hits the ground. Then solve the equation.
 Write an equation that can be solved to find when the potato is 40 feet off the ground. Then solve the equation.

The equation \(g(t) = 2 + 23.7t  4.9t^2\) models the height, in meters, of a pumpkin \(t\) seconds after it has been launched from a catapult.
 Is the pumpkin still in the air 8 seconds later? Explain or show how you know.
 At what value of \(t\) does the pumpkin hit the ground? Show your reasoning.
Student Response
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Anticipated Misconceptions
Watch for students who incorrectly substitute 40 for \(t\) and evaluate \(h(40)\), or who write \(g(t)=8\) instead of evaluating \(g(8)\). These students may be having trouble distinguishing between the input and output because the meaning of words such as “the height is a function of time” is still unclear to them. Remind these students that “height is a function of time” means that the variable in the formula given represents time, and that the formula can be used to find the height. Point out that \(h(t)\) represents the output for input \(t\).
Activity Synthesis
Make sure students understand what equations to write and what it means to solve each equation in the given contexts. Then, focus the discussion on how the solutions can be found, interpreted, and verified. Ask questions such as:
 “Once you have the equations, how did you find the solutions when the equation has a 0 on one side?” (By using the quadratic formula or by completing the square.)
 “What about when the equation has a nonzero number on one side?” (Rearrange it so that it is in standard form and has 0 on one side, and then use one of the solving methods.)
 “How many solutions did you get?” (Two for each equation.) “Do both solutions make sense in the given context?” (No. Only positive solutions are meaningful because they represent time after the objects were launched.)
 “How can we check if our solutions are accurate?” (Solve with a different method, substitute them back into the equation and see if the equation is true, or check by graphing.)
Consider displaying the graphs of the two functions to verify students’ solutions.
Design Principle(s): Optimize output (for explanation); Maximize metaawareness
17.3: Back to the Framer (15 minutes)
Activity
In this activity, students return to the framing problem they encountered when starting the unit. In that first lesson, they were challenged to use an entire sheet of paper to frame a picture and ensure that the thickness is uniform all around. At that time, students did not have adequate knowledge to solve the problem methodically, so they relied mainly on guessing and checking. Since then, students have developed their understandings around writing and solving quadratic equations. They are now ready to formulate the problem effectively and solve it methodically.
Students engage in aspects of mathematical modeling (MP4) as they identify the constraints in a situation, formulate a problem, construct a model, and interpret their solutions in context.
Because algebraic reasoning is the aim of this activity, graphing technology should not be used to solve the equations. It could be used to verify solutions during the activity synthesis, however.
Launch
Ask students to recall the framing problem from the beginning of the unit, how they tried to solve it, and what challenges they encountered. Consider preparing a copy of the picture and framing material from that lesson to serve as a visual aid. Tell students that they now have enough knowledge and skills to solve the problem more effectively and no longer have to rely on guessing and checking.
Arrange students in groups of 2. Give students a moment of quiet think time to interpret the equation \((7 + 2x)(4 + 2x) = 38\) in the first question and then discuss it with a partner. Display some sentence stems to help students articulate their interpretation. For example:
 The two factors in the equation represent . . .
 The number 38 represents . . .
 The variable \(x\) represents . . .
 Solving the equation means finding . . . . Pause afterward for a brief discussion. Make sure students understand the meaning of the equation before they proceed to solve it.
This activity was designed to be completed without graphing, so ask students to put away any graphing devices. Provide continued access to calculators for numerical computations.
If time is limited, consider asking students to complete only the first question.
Supports accessibility for: Conceptual processing; Language
Student Facing

In an earlier lesson, we tried to frame a picture that was 7 inches by 4 inches using an entire sheet of paper that was 4 inches by 2.5 inches. One equation we wrote was \((7+ 2x)(4 + 2x) = 38\).

Explain or show what the equation \((7+ 2x)(4 + 2x) = 38\) tells us about the situation and what it would mean to solve it. Use the diagram, as needed.
 Solve the equation without graphing. Show your reasoning.


Suppose you have another picture that is 10 inches by 5 inches, and are now using a fancy paper that is 8.5 inches by 4 inches to frame the picture. Again, the frame is to be uniform in thickness all the way around. No fancy framing paper is to be wasted!
Find out how thick the frame should be.
Student Response
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Student Facing
Are you ready for more?
Suppose that your border paper is 6 inches by 8 inches. You want to use all the paper to make a halfinch border around some rectangular picture.
 Find two possible pairs of length and width of a rectangular picture that could be framed with a halfinch border and no leftover materials.
 What must be true about the length and width of any rectangular picture that can be framed this way? Explain how you know.
Student Response
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Activity Synthesis
Invite students to share their solution strategies. If not mentioned in students’ explanations, make sure to discuss:
 Why negative solutions aren’t meaningful in this context.
 Which strategy they found effective and efficient for solving the equations. (The quadratic formula might not be the speediest way to solve the last equation, for example, as it can be solved by rewriting the equation in factored form and by completing the square.)
If time permits, consider asking students to verify their solution to the first question using the picture and framing materials from the blackline master from the first lesson of the unit. Ask students to cut the paper into strips that are as thick as the solution they calculated and arrange the strips around the picture. (If their solution is correct, there should be no leftover framing material and the frame should be uniform all around the picture.)
Lesson Synthesis
Lesson Synthesis
To help students consolidate the insights from this lesson, ask students to reflect on questions such as:
 “Before this lesson, you were able to solve many application problems that involve quadratic functions, but you didn’t use the quadratic formula. What does the quadratic formula allow you to do that you couldn’t do before?”
 “Suppose a situation can be modeled with a quadratic function, defined by an expression in factored form, and we want to find its zeros. We can write \((2x + 30)(\text4x +12)=0\) and solve the equation. Would you choose the quadratic formula to solve it? Why or why not?” (Probably not. The equation can be much more quickly solved by using the zero product property.)
 “What if we want to find when the function has a value of 75? Would the quadratic formula be a good way to solve \((2x + 30)(\text4x +12)=75\)? Why or why not?” (Yes. We cannot use the zero product property and completing the square is likely pretty laborious. We could graph, but the solutions might not be precise.)
 “How do we know what the \(a, b\), and \(c\) are in the equation \((2x + 30)(\text4x +12)=75\)? (We need to first rewrite it so that it is in the form of \(ax^2+bx+c=0\), and then identify those values.)
17.4: Cooldown  Tennis Ball Up, Tennis Ball Down (5 minutes)
CoolDown
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Student Lesson Summary
Student Facing
Quadratic equations that represent situations cannot always be neatly put into factored form or easily solved by finding square roots. Completing the square is a workable strategy, but for some equations, it may involve many cumbersome steps. Graphing is also a handy way to solve the equations, but it doesn’t always give us precise solutions.
With the quadratic formula, we can solve these equations more readily and precisely.
Here’s an example: Function \(h\) models the height of an object, in meters, \(t\) seconds after it is launched into the air. It is is defined by \(h(t)=\text5t^2+25t\).
To know how much time it would take the object to reach 15 meters, we could solve the equation \(15=\text5t^2+25t\). How should we do it?
 Rewriting it in standard form gives \(\text5t^2+25t15=0\). The expression on the left side of the equation cannot be written in factored form, however.
 Completing the square isn't convenient because the coefficient of the squared term is not a perfect square and the coefficient the linear term is an odd number.
 Let’s use the quadratic formula, using \(a=\text5,b=25,\text{ and }c=\text15\)!
\(\displaystyle \begin {align}\\t &=\dfrac{\textb \pm \sqrt{b^24ac}}{2a}\\ t &=\dfrac{\text25 \pm \sqrt{25^24(\text5)(\text15)}}{2(\text5)}\\ t &=\dfrac{\text25 \pm \sqrt{325}}{\text10} \end{align}\)
The expression \(\frac{\text25 \pm \sqrt{325}}{\text10}\) represents the two exact solutions of the equation.
We can also get approximate solutions by using a calculator, or by reasoning that \(\sqrt{325} \approx 18\).
The solutions tell us that there are two times after the launch when the object is at a height of 15 meters: at about 0.7 seconds (as the object is going up) and 4.3 seconds (as it comes back down).