Lesson 18
Applying the Quadratic Formula (Part 2)
18.1: Bits and Pieces (5 minutes)
Warmup
In this warmup, students evaluate variable expressions that resemble those in the quadratic formula. The aim is to preview the calculations that are the source of some common errors in solving equations using the quadratic formula, which students will analyze in the next activity.
Student Facing
Evaluate each expression for \(a=9\), \(b=\text5\), and \(c=\text2\)
 \(\textb\)
 \(b^2\)
 \(b^24ac\)
 \(\textb \pm \sqrt{a}\)
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Activity Synthesis
Invite students to share their responses and discuss any disagreement. For each expression, ask if they can think of an error someone might make when evaluating such an expression. Some possible errors:
 \(\textb\): Forgetting that \(\textb\) is really \(\text1(b)\) and the product of two negative numbers is positive.
 \(b^2\): When \(b\) is 5, evaluating \(\text(5)^2\), instead of \((\text5)^2\).
 \(b^24ac\): Forgetting that subtracting by \(4ac\) is equivalent to adding \(\text4ac\), or neglecting to see that if \(c\) is negative, \(\text4ac\) is positive, not negative.
 \(\textb \pm \sqrt{a}\): Neglecting the negative in front of \(b\), or neglecting to see that the expression takes two different values. Tell students that in the next activity, they will spot some errors in solving quadratic equations.
18.2: Using the Formula with Care (15 minutes)
Activity
Solving an equation with the quadratic formula involves multiple calculations, some of which may be prone to error. This activity acquaints students with some of the commonly made mistakes and encourages them to write and evaluate expressions carefully. As students analyze the progression of reasoning in worked solutions, they practice reasoning abstractly (MP2) and attending to precision in communicating a solution (MP3).
As students work, identify those who can clearly and correctly explain the errors in the worked solutions. Ask them to share their responses during the class discussion.
Launch
Arrange students in groups of 2. Give students a few minutes of quiet work time to solve 1–2 equations using the quadratic formula, and then a moment to discuss their response with their partner. Then, ask them to study the worked solutions (in the activity statement) for the same equations that they had solved and identify the errors.
If time permits, ask groups to identify the errors in the remaining equations. If they get stuck, suggest that they try solving the equations first and then look for the errors.
Provide access to calculators for numerical computations.
Student Facing
Here are four equations, followed by attempts to solve them using the quadratic formula. Each attempt contains at least one error.
 Solve 1–2 equations by using the quadratic formula.
 Then, find and describe the error(s) in the worked solutions of the same equations as the ones you solved.
Equation 1: \(\quad 2x^2 + 3 = 8x\)
Equation 2: \(\quad x^2 + 3x = 10\)
Equation 3: \(\quad 9x^22x1 = 0\)
Equation 4: \(\quad x^2  10x + 23 = 0\)
Here are the worked solutions with errors:
Equation 1: \(\quad 2x^2 + 3 = 8x\)
\(a=2,\, b= \text8,\, c=3\)
\(\displaystyle \begin {align} x&=\frac{\textb \pm \sqrt{b^24ac}}{2a} &\quad& \text { }\\x&=\frac{\text(\text8) \pm \sqrt{(\text8)^24(2)(3)}}{2(2)}\\ x&=\frac{8 \pm \sqrt{6424}}{4}\\x&=\frac{8 \pm \sqrt{40}}{4}\\x &=2 \pm \sqrt{10} \end {align}\)
Equation 2: \(\quad x^2 + 3x = 10\)
\(a = 1,\,b = 3,\, c = 10\)
\(\displaystyle \begin {align} x&=\frac{\textb \pm \sqrt{b^24ac}}{2a}\\ x&=\frac{\text3 \pm \sqrt{3^24(1)(10)}}{2(1)}\\ x&=\frac{\text3 \pm \sqrt{940}}{2}\\ x&=\frac{\text3 \pm \sqrt{\text31}}{2}\\ &\text{No solutions} \end{align}\)
Equation 3: \(\quad 9x^22x1=0\)
\(a = 9,\,b = \text2,\, c = \text1\)
\(\displaystyle \begin {align} x&=\frac{\textb \pm \sqrt{b^24ac}}{2a}\\ x&=\frac{2 \pm \sqrt{(\text2)^24(9)(\text1)}}{2}\\ x&=\frac{2 \pm \sqrt{4+36}}{2}\\x&=\frac{2 \pm \sqrt{40}}{2} \end{align}\)
Equation 4: \(\quad x^2  10x + 23 = 0\)
\(a = 1,\,b = \text10,\, c = 23\)
\(\displaystyle \begin {align} x&=\frac{\textb \pm \sqrt{b^24ac}}{2a}\\ x&=\frac{\text10 \pm \sqrt{(\text10)^24(1)(23)}}{2}\\ x&=\frac{\text10 \pm \sqrt{\text10092}}{2}\\ x&=\frac{\text10 \pm \sqrt{\text192}}{2}\\ &\text{No solutions} \end{align}\)
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Activity Synthesis
Display the worked solutions in the activity statement for all to see. Select previously identified students to identify and explain the error(s) in each worked solution.
After each student presents, ask the class to classify the error(s) by type and to explain their classification. For instance, are the errors careless mistakes or computational mistakes? Do they show gaps in understanding, incomplete communication, or a lack of precision?
If not mentioned by students, point out that these errors are easy to make but not always easy to notice. Sometimes, unless the solutions really make no sense, we won’t know if the solutions are actually correct or completely off unless we check them. Tell students that they will consider ways to check their solutions in the next activity.
Design Principle(s): Support sensemaking
Supports accessibility for: Visualspatial processing
18.3: Sure About That? (15 minutes)
Activity
Earlier, students practiced identifying likely mistakes in solving quadratic equations using the quadratic formula. This activity prompts students to check the solutions to quadratic equations and to recognize that there is more than one way to do so. Different approaches include substituting solutions to see if the equation is true, solving using a different algebraic method, or solving by graphing.
Monitor for students who verify the solutions to the equation \(2 + 30t  5t^2 = 47\) by:
 Solving the equation using another algebraic method and seeing if they get the same solutions. For instance, if they used the quadratic formula initially, they might choose to solve again by completing the square or by rewriting the equation in factored form.
 Substituting the solutions for \(t\) in the equation and seeing if the equation is true.
 Graphing \(y = 2 + 30t  5t^2\) and \(y= 47\) and seeing if the horizontal coordinates of the intersections match those of the solutions.
 Graphing \(y= \text45 + 30t  5t^2\) and checking the horizontal intercepts of the graph.
Identify students who use contrasting methods and ask them to share during discussion. The algebraic strategies are more precise but could lead to errors if the computations are performed incorrectly. The graphing strategies can be accomplished quickly with the use of technology but may only give approximate solutions.
Making calculators and graphing technology available gives students an opportunity to choose appropriate tools strategically (MP5).
Launch
Keep students in groups of 2. Ask students to think quietly about each question before conferring with their partner. Encourage partners to consider different ways of checking their solutions.
If time is limited, consider asking half of the class to answer the first question and the other half to answer the second question.
Provide access to calculators for numerical computations and to graphing technology, in case requested for solution checking.
Student Facing

The equation \(h(t) = 2 + 30t  5t^2\) represents the height, as a function of time, of a pumpkin that was catapulted up in the air. Height is measured in meters and time is measured in seconds.
 The pumpkin reached a maximum height of 47 meters. How many seconds after launch did that happen? Show your reasoning.
 Suppose someone was unconvinced by your solution. Find another way (besides the steps you already took) to show your solution is correct.

The equation \(r(p) = 80p  p^2\) models the revenue a band expects to collect as a function of the price of one concert ticket. Ticket prices and revenues are in dollars.
A band member says that a ticket price of either \$15.50 or \$74.50 would generate approximately \$1,000 in revenue. Do you agree? Show your reasoning.
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Student Facing
Are you ready for more?
Function \(g\) is defined by the equation \(g(t)=2+ 30t5t^2  47\). Its graph opens downward.
 Find the zeros of function \(g\) without graphing. Show your reasoning.
 Explain or show how the zeros you found can tell us the vertex of the graph of \(g\).
 Study the expressions that define functions \(g\) and \(h\) (which defined the height of the pumpkin). Explain how the maximum of function \(h\), once we know it, can tell us the maximum of \(g\).
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Activity Synthesis
Select previously identified students to present the different ways in which they checked their solutions. Sequence their presentations based on the precision of the verification (from more precise to less precise), as shown in the Activity Narrative. Highlight as many different strategies as time permits.
If no one mentions solving with a different strategy or graphing as ways to verify solutions, bring these up. Display a graph such as this one to show that the graph can immediately show that 74.50 is not a solution to the equation \(80p  p^2=1,\!000\), but 64.50 is very, very close.
Many students may have chosen to substitute the solutions back into the equation as a way to check them. Acknowledge that this is a very useful strategy for checking, but it also has its limits. Sometimes the solutions are not rational numbers—for example, \(x=2 \pm \frac12 \sqrt{10}\). Substituting such an expression back into the equation and evaluating it would mean complicated computations, which may themselves be subject to errors.
We could approximate the values of expressions like \(2 \pm \frac12 \sqrt{10}\) and write them as decimals before checking. This is a workable strategy if a calculator is handy.
Design Principle(s): Optimize output; Cultivate conversation
Lesson Synthesis
Lesson Synthesis
Invite students to name some familiar mistakes that are made when using the quadratic formula, not only those seen in the activities, but also mistakes that the students themselves might have made.
Then, to encourage students to think about ways of checking solutions, discuss questions such as:
 “What might be some good ways to check if \(x=\frac13\) and \(x=\frac52\) are solutions to \(6x^217x+5=0?\)” (Some possibilities:
 Substitute each solution into the equation and see if the equation is true. For example, see if \(6\left(\frac13\right)^2  17\left(\frac13\right) + 5\) equals 0.
 Rewrite \(6x^217x+5=0\) in factored form: \((3x1)(2x5)=0\) and apply the zero product property to solve.
 Graph \(y=6x^217x+5\) and find the \(x\)intercepts, when the graph has a \(y\)coordinate of 0.)
 “Can completing the square be used to check the solution?” (Yes, but it is not simple because the leading coefficient is not a perfect square and the linear coefficient is an odd number.)
 “Suppose we want to verify if \(x=1 \pm \sqrt3\) are correct solutions to \(x^2+2x2=0\). Is substituting the expressions in the equation an effective way to check?” (In this case, substituting \(1 \pm \sqrt3\) for \(x\) makes the checking cumbersome. The expression to be evaluated would look like: \((1+ \sqrt3)^2 + 2 (1 +\sqrt3) 2\), which is not easy to evaluate.)
 “What might be some good ways to check?” (We could:
 Solve by completing the square. The leading coefficient of \(x^2+2x2\) is 1 and the linear term coefficient is an even number, which makes it fairly simple to complete the square.
 Approximate the value of the solutions with a calculator, then graph \(y=x^2+2x2\) and see if the solutions match the \(x\)intercepts.)
18.4: Cooldown  Where Did It Go Wrong? (5 minutes)
CoolDown
Cooldowns for this lesson are available at one of our IM Certified Partners
Student Lesson Summary
Student Facing
The quadratic formula has many parts in it. A small error in any one part can lead to incorrect solutions.
Suppose we are solving \(2x^2 6 =11x\). To use the formula, let's rewrite it in the form of \(ax^2+bx+c=0\), which gives: \(2x^2 11x6=0\).
Here are some common errors to avoid:

Using the wrong values for \(a\), \(b\), and \(c\) in the formula.
\(x=\dfrac{\textb \pm \sqrt{b^24ac}}{2a}\)
\(x=\dfrac{\text11 \pm \sqrt{(\text11)^24(2)(\text6)}}{2(2)}\)
Nope! \(b\) is 11, so \(\textb\) is \(\text (\text11)\), which is 11, not 11.
\(x=\dfrac{11 \pm \sqrt{(\text11)^24(2)(\text6)}}{2(2)}\)
That’s better!

Forgetting to multiply 2 by \(a\) for the denominator in the formula.
\(x=\dfrac{11 \pm \sqrt{(\text11)^24(2)(\text6)}}{2}\)
Nope! The denominator is \(2a\), which is \(2(2)\) or 4.
\(x=\dfrac{11 \pm \sqrt{(\text11)^24(2)(\text6)}}{2(2)}\)
That’s better!

Forgetting that squaring a negative number produces a positive number.
\(x=\dfrac{11 \pm \sqrt{\text1214(2)(\text6)}}{4}\)
Nope! \((\text11)^2\) is 121, not 121.
\(x=\dfrac{11 \pm \sqrt{1214(2)(\text6)}}{4}\)
That’s better!

Forgetting that a negative number times a positive number is a negative number.
\(x=\dfrac{11 \pm \sqrt{12148}}{4}\)
Nope! \(4(2)(\text6)=\text48\) and \(121(\text48)\) is \(121+48\).
\(x=\dfrac{11 \pm \sqrt{121+48}}{4}\)
That’s better!

Making calculation errors or not following the properties of algebra.
\(x=\dfrac{11 \pm \sqrt{169}}{4}\)
\(x=11 \pm \sqrt{42.25}\)
Nope! Both parts of the numerator, the 11 and the \(\sqrt {169}\), get divided by 4. Also, \(\frac{\sqrt{169}}{4}\) is not \(\sqrt{42.25}\).
\(x=\dfrac{11 \pm 13}{4}\)
That’s better!
Let’s finish by evaluating \(\frac{11 \pm 13}{4}\) correctly:
\(\displaystyle \begin {align} x&= \dfrac{11 + 13}{4} \qquad&\text{or}& \qquad x= \dfrac{11  13}{4} \\ x &=\frac{24}{4} \qquad &\text{or}& \qquad x=\text\frac24 \\ x&=6 \qquad &\text{or}& \qquad x=\text\frac12 \end{align}\)
To make sure our solutions are indeed correct, we can substitute the solutions back into the original equations and see whether each solution keeps the equation true.
Checking 6 as a solution:
\(\displaystyle \begin{align}2x^2 6 &=11x\\ 2(6)^2 6 &=11(6)\\ 2(36) 6 &=66\\ 726 &= 66\\ 66 &=66 \end {align}\)
Checking \(\text\frac12\) as a solution:
\(\begin{align}2x^2 6 &=11x\\ 2 \left(\text\frac12\right)^2 6 &=11 \left(\text\frac12\right)\\ 2 \left(\frac14\right) 6 &=\text\frac{11}{2}\\ \frac12 6 &= \text5 \frac12\\ \text5\frac12 &=\text 5\frac12 \end {align}\)
We can also graph the equation \(y=2x^2 11x 6\) and find its \(x\)intercepts to see whether our solutions to \(2x^2 11x 6=0\) are accurate (or close to accurate).