9715b302-1b86-43bf-9793-bf22c13207bf.html

<h4>Launch</h4>
<p>Students may be tempted to solve these problems using Desmos or another technology as they did in the previous section. Remind students the purpose of this unit is writing quadratic equations from real solutions without graphing.</p>
<h4>Student Activity </h4>
<p>For questions 1-4, use the following information:</p>
<blockquote><p>A parabola crosses the \(x\)-axis at (-1,0) and (2,0). We want to write a quadratic equation with those \(x\)-intercepts. We can do this without graphing.</p></blockquote>
<ol class="os-raise-noindent">
  <li>When \(y=0\), what must \(x\) be?</li>
</ol>
<p><strong>Answer:</strong> We know that (-1,0) and (2,0) are points on the graph when \(y=0\). So there are two possibilities for \(x\): \(x=-1\) and \(x=2\).</p>
<p>\(x=-1\) and \(x=2\) are called the real solutions of the quadratic because they are the solutions of the equation when we set \(y=0\).</p>
<ol class="os-raise-noindent" start="2">
    <li>Rewrite the equations from number 1 so they both equal zero.</li>
</ol>
<p><strong>Answer:</strong> \(x+1=0\) and \(x-2=0\).</p>
<p>Because each of the equations equals zero, multiplying them together will also equal zero. </p>
<ol class="os-raise-noindent" start="3">
    <li> What equation do you get when you multiply the previous equations together? </li>
</ol>
<p><strong>Answer:</strong> \((x+1)(x-2)=0 \)</p>
<ol class="os-raise-noindent" start="4">
    <li> Multiply the binomials.</li>
</ol>
<p><strong>Answer:</strong> \(x^2-x-2=0\)</p>
<p>This answer can also give us the more general quadratic with those \(x\)-intercepts. We replace 0 with \(y\) to show that \(y\) varies and isn&rsquo;t always 0. So, \(y=x^2-x-2\) is a parabola that crosses the \(x\)-axis at -1 and 2.</p>

<h4>Activity Synthesis</h4>
<p>Students have already used factoring to help find the \(x\)-intercepts of a parabola. From the factorization of a quadratic, it is only one additional step to finding the real solutions of the equation. In this lesson, students will be working backward from real solutions to write a quadratic equation. Pointing out that they have already done this process in reverse may help some students.</p>

<p>Without more information or a graph, we don&rsquo;t discuss whether the associated parabola opens up or down, but this information could be added to an extension of the lesson where warranted.</p>

<p>Similarly, this lesson doesn&rsquo;t take into account a particular \(y\)-intercept. To extend this lesson, you could point out that any quadratic of the form \(y=a(x-x\)<sub>1</sub>\()(x-x\)<sub>2</sub>\()\), where \(x\)<sub>1</sub> and \(x\)<sub>2</sub> are the real roots, will be a quadratic that meets the requirements. Then, they can plug in (0, y) for the \(y\)-intercept, or even any known point on the parabola to find the particular value for \(a\) that fits the additional point.</p>
<br>
<h3>7.13.4: Self Check</h3>
<!--BEGIN SELF CHECK INTRO BEFORE Tables -->   

<p class="os-raise-text-bold"><em>After the activity, students will answer the following question to check their understanding of the concepts explored in the activity.</em></p>

<p class="os-raise-text-bold">QUESTION:</p>
<p>Write a quadratic from the real solutions \(x=-3\) and \(x=4\) without graphing.</p>

<!--SELF CHECK table-->
<table class="os-raise-textheavytable">
   <thead>
    <tr>
      <th scope="col">Answers</th>
      <th scope="col">Feedback</th>
    </tr>
 </thead>
  <tbody>
    <tr>
      <td>
        <p><strong>Answers</strong></p>
      </td>
      <td>
        <p><strong>Feedback</strong></p>
      </td>
    </tr>
    <tr>
      <td>
        <p>\(y=x^2-x+12\)</p>
      </td>
      <td>
        <p>Incorrect. Let&rsquo;s try again a different way: \(x=-3\) and \(x=4\) becomes \(x+3=0\) and \(x-4=0\). \(y=(x+3)(x-4)= x^2-x-12\). Check your signs when you use FOIL to multiply the binomials.</p>
      </td>
    </tr>
    <tr>
      <td>
        <p>\(y=x^2-x-12\)</p>
      </td>
      <td>
        <p>That&rsquo;s correct! Check yourself: \(x=-3\) and \(x=4\) becomes \(x+3=0\) and \(x-4=0\). \(y=(x+3)(x-4)= x^2-x-12\).
          </p>
      </td>
    </tr>
    <tr>
      <td>
        <p>\(y=x^2+x-12\)</p>
      </td>
      <td>
        <p>Incorrect. Let&rsquo;s try again a different way: Remember that \(x=-3\) and \(x=4\) becomes \(x+3=0\) and \(x-4=0\). Then \(y=(x+3)(x-4)= x^2-x-12\). </p>
      </td>
    </tr>
    <tr>
      <td>
        <p>\(y=x^2+x+12\)</p>
      </td>
      <td>
        <p>Incorrect. Let&rsquo;s try again a different way: \(x=-3\) and \(x=4\) becomes \(x+3=0\) and \(x-4=0\). \(y=(x+3)(x-4)= x^2-x-12\). Check your signs when you use FOIL to multiply the binomials.</p>
      </td>
    </tr>
    </tbody>
  </table>
<br>
<!--END SELF CHECK INTRO BEFORE Tables -->   
<br>
<h3>7.13.4: Additional Resources</h3>
<em><strong>
    <p>The following content is available to students who would like more support based on their experience with the self check. Students will not automatically have access to this content, so you may wish to share it with those who could benefit from it.</p>
  </strong></em>
<p>When you write a quadratic in factored form, that helps you find the \(x\)-intercepts.<br> 
\(y=x^2+2x-3\)<br>
\(y=(x+3)(x-1)\)<br>
</p>
<p>When we set \(y=0\), we find the places where the parabola crosses the \(x\)-axis.</p>

<p>\(0=(x+3)(x-1)\)</p>

<p>So, \(0=x+3\) and \(0=x-1\)<br>
\(x=-3\) and 1. </p>

<p>-3 and 1 are called the real solutions to the quadratic equation.</p>

<p>We can also start with the real solutions and use them to write quadratic equations.</p>

<p>When we know that \(x=-3\) and 1, then that means \(0=x+3\) and \(0=x-1\).</p>

<p>So, \(0=(x+3)(x-1)\).</p>

<p>And, using the FOIL method to multiply this out, we come to <br>
\(0=x^2+2x-3.\)</p>

<p>When \(y\) is not equal to zero, we have the following quadratic equation</p>

<p>\(y=x^2+2x-3\), which is exactly where we started!</p>


<h4>Try It: Writing Quadratic Equations from Real Solutions</h4>
<p>Write a quadratic from the real solutions \(x=1\) and \(x=7\) without graphing.</p>
<p><strong>Answer:</strong> \(x^2-8x+7\)</p>
<p>\(x=1\) and \(x=7\) become \(x-1=0\) and \(x-7=0\). \(y=(x-1)(x-7)= x^2-x-7x+7= x^2-8x+7\).</p>