99d4617e-97e2-4728-9037-0d01792f559a.html

<h3>Cool Down (10 minutes)</h3>
<p>This cool down activity aims to uncover some common misconceptions in solving quadratic equations and to reinforce that certain familiar moves for solving equations are not effective. Students critique several arguments on how to solve quadratic equations. In articulating why certain lines of reasoning are correct or incorrect, they practice constructing logical arguments.</p>
<p>As students work, look for students who:</p>
<ul class="os-raise-noindent">
  <li> explain both the error in Priya's argument and the validity of Mai's argument in terms of the zero product property.</li>
  <li> notice that Diego's method disregards the second solution of the equation.</li>
  <li> create and use a graph to verify their critique of Priya, Mai, or Diego's work (for example, graphing to show that \(x^2&minus;10x\) has two \(x\)-intercepts).</li>
</ul>
<p>Multiplying or dividing both sides of an equation by a variable expression can change the solution set of an equation, either by eliminating a solution (as shown in Diego's method) or by introducing a new solution (for example, starting with the equation \(x=3\) and multiplying both sides by \(x\) to get \(x^2=3x\) gives an equation that now has two solutions). Students will learn more about such moves in a later course.</p>
<h4>Launch</h4>
<p>Keep students in groups of two and ask them to work quietly on the questions before discussing their responses with a partner.</p>
<div class="os-raise-extrasupport">
  <div class="os-raise-extrasupport-header">
    <p class="os-raise-extrasupport-title">Support for English Language Learners</p>
  </div>
  <div class="os-raise-extrasupport-body">
    <p class="os-raise-extrasupport-name">MLR 1 Discussion Supports: Writing, Conversing</p>
    <p>Use this routine to help students improve their written responses by providing them with multiple opportunities to clarify their explanations through conversation. At the appropriate time, give students time to meet with two to three partners to share their response to the first question. Students should first check to see if they agree with each other about Priya’s reasoning. Provide listeners with prompts for feedback that will help their partner add detail to strengthen and clarify their ideas. For example, "Your explanation tells me . . .", "Can you say more about why you . . . ?", and "A detail (or word) you could add is _____, because . . . ." Give students three to four minutes to revise their initial draft based on feedback from their peers. This will help students evaluate the written mathematical arguments of others and improve their own written responses about solving quadratic equations.</p>
    <p class="os-raise-text-italicize">Design Principle(s): Optimize output (for justification); Cultivate conversation</p>
    <p class="os-raise-extrasupport-title">Provide support for students</p>
      <p>
        <a href="https://k12.openstax.org/contents/raise/resources/a5ae5bd09b27a5f53239a539c6009c19c92f7db7" target="_blank">Distribute graphic organizers</a>
           to the students to assist them with participating in this routine.
      </p>
  </div>
</div>

<br>
<h4>Student Activity</h4>
<ol class="os-raise-noindent">
  <li> Consider \((x&minus;5)(x +1)=7\). Priya reasons that if this is true, then either \(x&minus;5=7\) or \(x+1=7\). So, the solutions to the original equation are 12 and 6.</li>
</ol>
<p>Do you agree? If not, where was the mistake in Priya's reasoning?</p>
<p><strong>Answer:</strong> Disagree. Priya solved the equation using the reasoning we would use with the zero product property, but the zero product property only works if the product of two factors is 0. We can tell that 12 isn't a solution because \((12&minus;5)(12+1)\) is 91, not 7.</p>
<ol class="os-raise-noindent" start="2">
  <li>Consider \(x^2&minus;10x=0\). Diego says that to solve we can just divide each side by \(x\) to get \(x&minus;10=0\), so the solution is 10. Mai says, "I wrote the expression on the left in factored form, which gives \(x(x&minus;10)=0\), and ended up with two solutions: 0 and 10."</li>
</ol>
<p>Do you agree with either strategy? Be prepared to show your reasoning.</p>
<p><strong>Answer: </strong>Agree with Mai's strategy. By substituting 0 for \(x\) and then 10 for \(x\), you can see that they are both solutions to the original equation. Diego's strategy of dividing by a variable eliminates one of the solutions, leaving only one solution.</p>
<ol class="os-raise-noindent" start="3">
  <li>Consider the equation \(x^2=-16\). Mona says that there are two solutions, \(x=4\) or \(x=-4\).&nbsp;</li>
</ol>
<p>Do you agree? If not, where was the mistake in Mona's reasoning?</p>
<p><strong>Answer:</strong> Disagree. There are no numbers we can square to get a negative product. The equation has no solutions.</p>
<ol class="os-raise-noindent" start="4">
  <li>Consider the equation \((x&minus;3)(x&minus;3)=0\). Dominic says that there are two solutions, \(x=3\) or \(x=-3\).&nbsp;</li>
</ol>
<p>Do you agree? If not, where was the mistake in Dominic's reasoning?</p>
<p><strong>Answer:</strong> Disagree. Only 3 makes the equation true. The solution is \(x=3\).</p>
<h4>Activity Synthesis</h4>
<p>Select previously identified students to share their responses and reasoning. Here are some key observations to highlight:</p>
<ul class="os-raise-noindent">
  <li> For the first question, make sure students understand that the zero product property only works when the product of the factors is 0. While substituting 6 for \(x\) in the expression does produce 7, substituting 12 does not give the same result. </li>
  <li> For the second question, consider graphing the function \(y=x^2&minus;10x\) so students can see that the graph intersects the \(x\)-axis at two points, which means that there are two \(x\)-values that give a zero output: 0 and 10. Rewriting \(x^2&minus;10x\) into the factored form, writing it to equal 0, and solving \(x(x&minus;10)=0\) allow us to see that this is indeed the case. </li>
  <li> Dividing each side by a variable (as Diego did) seems to enable us to isolate the remaining variable, but only one solution remains. Dividing each side of an equation by \(x\) is not a valid move because when \(x\) is 0, the expressions on each side become undefined. </li>
</ul>