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<h4>Activity (10 minutes)</h4>
<p>This is the first of two activities that help students derive the quadratic formula. Here students try to complete the square for a quadratic equation but without evaluating some of the numerical expressions along the way. Doing so enables them to recognize the parts of the quadratic formula but in numerical form (for example, instead of seeing \(-b \pm \sqrt{b^2-4ac}\), they see \(-5 \pm \sqrt{5^2-12}\)).</p>
<p>The linear coefficient of the given equation is an odd number. Students learn that there are strategies to transform the quadratic expression such that it has an even number for the linear coefficient and 1 for the leading coefficient, which makes it much easier to complete the square. These strategies involve multiplying the equation by a helpful factor and temporarily using simpler variables to stand in for complicated parts of an expression.</p>
<p>(The strategy of multiplying an equation by a number to make the \(a\) a perfect square is illustrated in an extension activity in the last lesson on completing the square. Students who completed that activity may recall this approach. Using a simple variable to stand in for a messier expression is also explored in an extension activity in an earlier lesson-the last lesson on rewriting expressions in factored form.)</p>
<h4>Launch</h4>
<p>The quadratic equation \(x^2+5x+3=0\) is in the form of \(ax^2+bx+c=0\). What are the values of \(a\), \(b\), and \(c\)? Talk to a partner about whether it would be relatively simple to complete the square from this equation. (It's doable, but probably pretty messy, as \(b\) is an odd number.)</p>
<p>Tell students that one way to make it easier to complete the square is to multiply the equation by a number such that the \(b\) is an even number and the \(a\) is a perfect square.</p>
<ul class="os-raise-noindent">
<li> Let's try multiplying \(x^2+5x+3=0\) by 2. We get \(2x^2+10x+6=0\). (The right side remains 0 because 0 times any number is 0.) We have 10, an even number for \(b\), but the \(a\) is not a perfect square. </li>
<li> Let's try multiplying it by 4. We get \(4x^2+20x+12=0\). Now the \(a\) is a perfect square and the \(b\) is still an even number. To complete the square, it helps to isolate the existing constant and rewrite it as \(4x^2+10x=-12\). </li>
<li> It's still not immediately obvious what constant term to add to make a perfect square. It would be easier if the \(a\) or the coefficient of the squared term is 1. One way to deal with this is to think of \(4x^2\) as (something)\(^2\). What would the "something" be? \((2x)\) </li>
<li> If we use a placeholder \(P\) to stand for \(2x\), we can write: </li>
</ul>
<p class="os-raise-noindent">\(\begin{array}{rcl}4x^2+20x&=&-12\\{(2x)}^2+10(2x)&=&-12\\{(P)}^2+10(P)&=&-12\end{array}\)</p>
<p>Ask students to try completing the square for this simpler-looking equation.</p>
<h4>Student Activity</h4>
<ol class="os-raise-noindent">
<li>One way to solve the quadratic equation \(x^2+5x+3=0\) is by completing the square. A partially solved equation is shown here. Study Steps 1–4.</li>
</ol>
<p>Original equation<br>
\(x^2+5x+3=0\)</p>
<p><strong>Step 1</strong> - Multiply each side by 4. This helps us avoid some tricky fractions. (In general, multiply each side of the equation by 4 times the coefficient of the \(x^2\) term. In this case, that is just 4 times 1, which is 4.)<br>
\(4x^2+20x+12=0\)</p>
<p><strong>Step 2</strong> - Subtract 12 from each side.<br>
\(4x^2+20x=-12\)</p>
<p><strong>Step 3</strong> - Rewrite \(4x^2\) as \((2x)^2\) and \(20x\) as \(10(2x)\).<br>
\((2x)^2+10(2x)=-12\)</p>
<p><strong>Step 4</strong> - Use \(P\) as a placeholder for \(2x\).<br>
\(P^2+10P=-12\)</p>
<p>For steps 5–11, let \(P\) serve as a placeholder for \(2x\), continue to solve for \( x\) but without evaluating any part of the expression. For each step, explain the mathematical operation(s) being performed.</p>
<p><strong>Step 5</strong> - \(\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\)<br>
\(P^2+10P+\_\_\_\_^2=-12+\_\_\_\_^2\)</p>
<p><strong>Step 6</strong> - \(\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\)<br>
\((P+\_\_\_\_)^2=-12+\_\_\_\_^2\)</p>
<p><strong>Step 7</strong> - \(\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\)<br>
\( \sqrt{(P+\_\_\_\_)^2}=\sqrt{-12+\_\_\_\_^2}\)</p>
<p>\(P+\_\_\_\_= \pm \sqrt{-12+\_\_\_\_^2}\)</p>
<p><strong>Step 8</strong> - \(\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\)<br>
\(P=\_\_\_\_ \pm \sqrt{-12+\_\_\_\_^2}\)</p>
<p><strong>Step 9</strong> - \(\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\)<br>
\(P=\_\_\_\_ \pm \sqrt{\_\_\_\_^2-12}\)</p>
<p>Step 10<br>
\(\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\)<br>
\(2x=\_\_\_\_ \pm \sqrt {\_\_\_\_^2-12}\)</p>
<p>Step 11<br>
\(\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\)<br>\(x=\)</p>
<p>Write down your answer, then select the <strong>solution</strong> button to compare your work.</p>
<p><strong>Answer:</strong><br>
Step 5<br>
Add to complete the square.<br>
\(P^2+10P+5^2=-12+5^2\)</p>
<p><strong>Step 6</strong> - Write the left side as a squared factor.<br>
\((P+5)^2=-12+5^2\)</p>
<p><strong>Step 7</strong> - Find the square root of both sides of the equation.
<br>
\( \sqrt {(P+5)^2}=\sqrt {-12+5^2}\)<br>
\(P+5= \pm \sqrt {-12+5^2}\)
</p>
<p><strong>Step 8</strong> - Subtract 5 from both sides to isolate.<br>
\(P=-5 \pm \sqrt {-12+5^2}\)</p>
<p><strong>Step 9</strong> - Rearrange the expression under the square root sign.<br>
\(P=-5 \pm \sqrt {5^2-12}\)</p>
<p>Step 10<br>
Re-substitute \(2x\) for \(P\).<br>
\(2x=-5 \pm \sqrt {5^2-12}\)</p>
<p>Step 11<br>
Divide each side by 2 to isolate \(x\).<br>
\(x=\frac{-5 \pm \sqrt{5^2-12}}{2}\) </p>
<ol class="os-raise-noindent" start="2">
<li>Explain how the solution is related to the quadratic formula.</li>
</ol>
<p>Write down your answer, then select the <strong>solution</strong> button to compare your work.</p>
<p><strong>Answer:</strong> The numbers in the solution are the values of \(a\), \(b\), and \(c\) in the original equation, \(x^2+5x+3=0\), which is in the form of \(ax^2+bx+c=0\). The -5 is \(-b\). The \(5^2\) is \(b^2\). The 12, which is \(4(3)(1)\), is the value of \(4ac\). The \(a\) in the original equation is 1, and \(2(1)\) is 2, which is the number in the denominator in the solution.</p>
<h4>Activity Synthesis</h4>
<p>Highlight the connections between the numbers in the solution \(x=\frac{-5 \pm \sqrt{5^2-12}}{2}\) and the parameters in the original equation.</p>
<ul class="os-raise-noindent">
<li> The -5 in the expression is \(-b\). </li>
<li> The \(5^2\) is \(b^2\). </li>
<li> The 12 is \(4(1)(3)\) or \(4ac\). </li>
<li> The 2 in the denominator is \(2(1)\) or \(2a\). </li>
</ul>
<h3>9.8.1: 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>
<!--SELF CHECK QUESTION GOES BEFORE THE Table -->
<p class="os-raise-text-bold">QUESTION:</p>
<p>After completing the square, Odem used the following steps when deriving the quadratic formula.</p>
<p>\(x^2+8x+16= -11+16\)<br>
\(x^2+8x+16=5\)</p>
<p>What should he do next?</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>Subtract 16 from both sides.</p>
</td>
<td>
<p>Incorrect. Let's try again a different way: The 16 was just added in to create the perfect square. The answer is to factor the perfect square trinomial.</p>
</td>
</tr>
<tr>
<td>
<p>Take the square root of both sides.</p>
</td>
<td>
<p>Incorrect. Let's try again a different way: The trinomial must be factored before taking the square root. The answer is to factor the perfect square trinomial.</p>
</td>
</tr>
<tr>
<td>
<p>Factor the perfect square trinomial.</p>
</td>
<td>
<p>That's correct! Check yourself: After completing the square and adding the new value to both sides, factor the trinomial.</p>
</td>
</tr>
<tr>
<td>
<p>Complete the square again.</p>
</td>
<td>
<p>Incorrect. Let's try again a different way: The square only needs to be completed once to create a perfect square trinomial. The answer is to factor the perfect square trinomial.</p>
</td>
</tr>
</tbody>
</table>
<br>
<!--END SELF CHECK INTRO BEFORE Tables -->
<br>
<h3>9.8.1: 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>
<h4>Deriving the Quadratic Formula Using an Example</h4>
<p>Here is another example that goes through the process of deriving the quadratic formula:</p>
<p>Original equation<br>
\(x^2+7x+4=0\)</p>
<p><strong>Step 1</strong> - Multiply each side by 4. This helps us avoid some tricky fractions. (In general, multiply each side of the equation by 4 times the coefficient of the \(x^2\) term. In this case, that is just 4 times 1, which is 4.)<br>
\(4x^2+28x+16=0\)</p>
<p><strong>Step 2</strong> - Subtract 16 from each side.<br>
\(4x^2+28x=-16\)</p>
<p><strong>Step 3</strong> - Rewrite.<br>
\(4x^2\) as \((2x)^2\) and \(28x\) as \(14(2x)\)<br>
\((2x)^2+14(2x)=-16\)</p>
<p><strong>Step 4</strong> - Use \(P\) as a placeholder for \(2x\).<br>
\(P^2+14P=-16\)</p>
<p><strong>Step 5</strong> - Add to complete the square.<br>
\(P^2+14P+7^2=-16+7^2\)</p>
<p><strong>Step 6</strong> - Write the left side as a squared factor.<br>
\((P+7)^2=-16+7^2\)</p>
<p><strong>Step 7</strong> - Find the square root of the expression on the right.<br>
\(P+7= \pm \sqrt{-16+7^2}\)</p>
<p><strong>Step 8</strong> - Subtract 7 from both sides to isolate.<br>
\(P=-7 \pm \sqrt{-16+7^2}\)</p>
<p><strong>Step 9</strong> - Rearrange the expression under the square root sign.<br>
\(P=-7 \pm \sqrt{7^2-16}\)</p>
<p>Step 10<br>
Re-substitute \(2x\) for \(P\).<br>
\(2x=-7 \pm \sqrt{7^2-16}\)</p>
<p>Step 11<br>
Divide each side by 2 to isolate \(x\).<br>
\(x=\frac{-7 \pm \sqrt{7^2-16}}{2}\) </p>
<p>Notice the original equation, \(y=x^2+7x+4\), has \(a=1\), \(b=7\), \(c=4\), and the last step is of the form \(x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\), which is the quadratic formula.</p>
<h4>Try It: Deriving the Quadratic Formula Using an Example</h4>
<p>Jaiden was deriving the quadratic formula from the quadratic \(x^2+11x+5=0\). What is the step after the step shown below?</p>
<p>\((x+\frac{11}{2})^2=\frac{11^2-4(1)(5)}{4(1)}\)</p>
<p>\((x+\frac{11}{2})^2=\frac{121-20}{4}\)</p>
<p>Write down your answer, then select the <strong>solution</strong> button to compare your work.</p>
<h5>Solution</h5>
<p>Here is how to find the next step:</p>
<p>To isolate \(x\), take the square root of both sides:</p>
<p>\(\sqrt{(x+\frac{11}{2})^2}=\sqrt{\frac{121-20}{4}}\)</p>