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<h4>Activity (15 minutes)</h4>
<p>In the previous activity, students solved a quadratic equation by completing the square, but they did so without evaluating any of the numerical expressions. Students explained each step along the solving process and saw how it produced a solution that looks almost exactly like the quadratic formula, except that it contains numbers instead of \(a\), \(b\), and \(c\). That activity gave students concrete insights into the process of deriving the quadratic formula, preparing them to do the same in more abstract terms.</p>
<p>In this activity, students study a series of steps taken to solve \(ax^2+bx+c=0\) by completing the square and make sense of how it leads to the quadratic formula. Along the way, they see that the solving process involves similar maneuvers as those seen earlier, for example, multiplying the equation by a factor to make the leading coefficient a perfect square, writing the linear coefficient as 2 times two numbers, and adding the square of one of the numbers to complete the square. They see that the result of solving the equation by completing the square is the quadratic formula.</p>
<h4>Launch</h4>
<p>Tell students that they will now study a worked-out solution to \(ax^2+bx+c=0\). There are no numbers in this equation, but the process of solving should be familiar. Ask students to analyze the solution and record an explanation for each step. (They should explain why each step is taken, not only what happens in each step.)</p>
<br>
<div class="os-raise-extrasupport">
<div class="os-raise-extrasupport-header">
<p class="os-raise-extrasupport-title">Support for Students with Disabilities</p>
<p class="os-raise-extrasupport-name">Representation: Internalize Comprehension</p>
</div>
<div class="os-raise-extrasupport-body">
<p>Use color and annotations to support information processing. Invite students to use color to identify and keep track of terms from one step to the next.</p>
<p class="os-raise-text-italicize">Supports accessibility for: Visual-spatial processing; Organization</p>
</div>
</div>
<br>
<h4>Student Activity</h4>
<p>Here is one way to make sense of how the quadratic formula came about. Study the derivation until you can explain what happened in each step. Write down your explanation for each step.</p>
<p>Original equation<br>
\(ax^2+bx+c=0\)</p>
<p><strong>Step 1</strong> - \(4a^2x^2+4abx+4ac=0\)</p>
<p><strong>Step 2</strong> - \(4a^2x^2+4abx=-4ac\)</p>
<p><strong>Step 3</strong> - \((2ax)^2+2b(2ax)=-4ac\)</p>
<p><strong>Step 4</strong> - \(M^2+2bM=-4ac\)</p>
<p><strong>Step 5</strong> - \(M^2+2bM+b^2=-4ac+b^2\)</p>
<p><strong>Step 6</strong> - \((M+b)^2=b^2-4ac\)</p>
<p><strong>Step 7</strong> - \( \sqrt{(M+b)^2} =\sqrt{b^2-4ac}\)</p>
<p>\(M+b= \pm \sqrt{b^2-4ac}\)</p>
<p><strong>Step 8</strong> - \(M=-b \pm \sqrt{b^2-4ac}\)</p>
<p><strong>Step 9</strong> - \(2ax=-b \pm \sqrt{b^2-4ac}\)</p>
<p><strong>Step 10</strong> - <br>
\(x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\)</p>
<p>Write down your answer, then select the <strong>solution</strong> button to compare your work.</p>
<p><strong>Answer:</strong><br>
Original equation<br>
\(ax^2+bx+c=0\)</p>
<p><strong>Step 1</strong> - Multiply by \(4a\) so the coefficient of \(x^2\) is a perfect square.<br>
\(4a^2x^2+4abx+4ac=0\)</p>
<p><strong>Step 2</strong> - Isolate the constant term to the right.<br>
\(4a^2x^2+4abx=-4ac\)</p>
<p><strong>Step 3</strong> - Write the squared term as (something)\(^2\).<br>
Write the linear term as \(2b \;\cdot\) (something).<br>
\((2ax)^2+2b(2ax)=-4ac\)</p>
<p><strong>Step 4</strong> - Use \(M\) as the placeholder for \(2ax\).<br>
\(M^2+2bM=-4ac\)</p>
<p><strong>Step 5</strong> - Add \(b^2\) to complete the square.<br>
\(M^2+2bM+b^2=-4ac+b^2\)</p>
<p><strong>Step 6</strong> - Write the left side as a squared factor and commute the terms on the right.<br>
\((M+b)^2=b^2-4ac\)</p>
<p><strong>Step 7</strong> - Find the square roots of both sides of the equation.<br>
\( \sqrt{(M+b)^2} =\sqrt{b^2-4ac}\)</p>
<p>\(M+b= \pm \sqrt{b^2-4ac}\)</p>
<p><strong>Step 8</strong> - Subtract \(b\) to isolate \(M\).<br>
\(M=-b \pm \sqrt{b^2-4ac}\)</p>
<p><strong>Step 9</strong> - Replace \(M\) with \(2ax\).<br>
\(2ax=-b \pm \sqrt{b^2-4ac}\)</p>
<p><p><strong>Step 10</strong> - <br>
Divide both sides by \(2a\) to isolate \(x\).<br>
\(x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)</p>
<h4>Student Facing Extension</h4>
<h5>Are you ready for more?</h5>
<p>Here is another way to derive the quadratic formula by completing the square.</p>
<ul class="os-raise-noindent">
<li> First, divide each side of the equation \(ax^2+bx+c=0\) by \(a\) to get \(x^2+\frac{b}{a}x+\frac{c}{a}=0\). </li>
<li> Then, complete the square for \(x^2+\frac{b}{a}x+\frac{c}{a}=0\). </li>
</ul>
<p>In problems 1–10, briefly explain what happens in each step by answering the question. Use the "^" symbol to enter an exponent.</p>
<p>Original equation<br>
\(x^2+\frac{b}{a}x+\frac{c}{a}=0\)</p>
<ol class="os-raise-noindent">
<li>Step 1<br>
What constant should be subtracted from both sides?</li>
</ol>
<p><strong>Answer:</strong> \(-\frac{c}{a}\)</p>
<p>Now the equation looks like</p>
<p>\(x^2+\frac{b}{a}x= -\frac{c}{a}\).</p>
<ol class="os-raise-noindent" start="2">
<li>Step 2<br>
What constant do we need to add to both sides to complete the square?</li>
</ol>
<p><strong>Answer:</strong> \((\frac{b}{2a})^2\)</p>
<p>To complete the square, we add it to both sides. Now the equation looks like</p>
<p>\(x^2+(\frac{b}{2a})x+(\frac{b}{2a})^2= -\frac{c}{a}+(\frac{b}{2a})^2\).</p>
<p>Notice that the denominators are different on the left side of the equation (for the second and third terms). In order to form a common denominator of \(2a\), we must multiply the middle term by a factor of \frac{2}{2}.</p>
<p>\(x^2+\frac{2}{2}(\frac{b}{a})x+ (\frac{b}{2a})^2= -\frac{c}{a}+(\frac{b}{2a})^2\).</p>
<p>Regrouping the factors in the middle term differently yields
</p>
<p>\(x^2+2(\frac{b}{2a})x+ (\frac{b}{2a})^2= -\frac{c}{a}+(\frac{b}{2a})^2\).</p>
<ol class="os-raise-noindent" start="3">
<li>Step 3<br>
Now we can complete the square. What is the binomial that is squared on the left side?</li>
</ol>
<p><strong>Answer:</strong> \(x+\frac{b}{2a}\)</p>
<p>Now the equation looks like<br>
\((x+\frac{b}{2a})^2= -\frac{c}{a}+\frac{b^2}{4a^2}\).</p>
<ol class="os-raise-noindent" start="4">
<li>Step 4<br>
Next, we add the fractions on the right side of the equation. What will the common denominator be?</li>
</ol>
<p><strong>Answer:</strong> The common denominator will be \(4a^2\).</p>
<p>Now the equation looks like <br>
\((x+\frac{b}{2a})^2= -\frac{4ac}{4a^2}+\frac{b^2}{4a^2}\).</p>
<ol class="os-raise-noindent" start="5">
<li>Step 5<br>
Once the fractions on the right side of the equation are added together, what will the numerator of that new fraction be?</li>
</ol>
<p><strong>Answer:</strong> \(b^2-4ac\)</p>
<p>Now the equation looks like<br>
\((x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}\).</p>
<ol class="os-raise-noindent" start="6">
<li>Step 6<br>
What is the next step necessary to isolate \(x\) on the left side?</li>
</ol>
<p><strong>Answer:</strong> Take the square root of both sides.</p>
<p>Remember to put the \(\pm\) in front of the radical! Now the equation looks like<br>
\(x+\frac{b}{2a}= \pm \sqrt{\frac{b^2-4ac}{4a^2}}\).</p>
<ol class="os-raise-noindent" start="7">
<li>Step 7<br>
How can the square root be simplified?</li>
</ol>
<p><strong>Answer:</strong> The square root of a fraction is the square root of the numerator divided by the square root of the denominator.</p>
<p>\(x+\frac b{2a}=\pm\frac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}\)</p>
<ol class="os-raise-noindent" start="8">
<li>Step 8<br>
What is the denominator of the fraction on the right side after simplification?</li>
</ol>
<p><strong>Answer:</strong> \(2a\)</p>
<p>Now the equation looks like<br>
\(x+\frac b{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}\).</p>
<ol class="os-raise-noindent" start="9">
<li>Step 9<br>
What is the next step I need to take to isolate \(x\) on the left side of the equation?</li>
</ol>
<p><strong>Answer:</strong> Subtract \(\frac{b}{2a}\) from both sides.</p>
<p>Now the equation looks like<br>
\(x=-\frac b{2a} \pm\frac{\sqrt{b^2-4ac}}{2a}\).</p>
<ol class="os-raise-noindent" start="10">
<li><strong>Step 10</strong> - <br>
What do you notice about the denominator of both fractions on the right side?</li>
</ol>
<p><strong>Answer:</strong> They are the same.</p>
<p>Add the fractions, and we come to the result: the quadratic equation,<br>
\(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\).</p>
<h4>Video: Deriving the Quadratic Formula</h4>
<p>Watch the following video to learn more about deriving the quadratic formula.</p>
<div class="os-raise-d-flex-nowrap os-raise-justify-content-center">
<div class="os-raise-video-container"><video controls="true" crossorigin="anonymous">
<source src="https://k12.openstax.org/contents/raise/resources/ea0b4447ec781db56cae7dec10ddfe3cace6df40">
<track default="true" kind="captions" label="On" src="https://k12.openstax.org/contents/raise/resources/5c46d09e7ba2332487e618cbe92d9c7796441239" srclang="en_us">https://k12.openstax.org/contents/raise/resources/cdad89b7a5ee8c0eacfcbd8a970217904f3ade81
</video></div>
</div>
<br>
<br>
<h4>Anticipated Misconceptions</h4>
<p>Some students may think that we multiply just by 4 and wonder where the \(a\) comes from. Point out that we need the first term to be a perfect square. In the previous activity, the \(a\) value was 1 (which was already a perfect square). We can't be certain that \(a\) is a perfect square, so we multiply by 4 and \(a\) to make the first term \(4a^2x^2\), which is \((2ax)^2\), a perfect square.</p>
<p>Some may wonder where the \(M\) comes from. Consider displaying a completed solution from the previous activity that parallels this work. Remind students of the placeholder \(P\) used there. Invite them to compare other steps in the previous activity to those in this activity to help them explain the derivation.</p>
<h4>Activity Synthesis</h4>
<p>Invite students to share their explanations for each step. Highlight a few key maneuvers:</p>
<ul class="os-raise-noindent">
<li> Multiplying the equation by \(4a\) makes the coefficient of \(x^2\) a perfect square and the coefficient of the linear term an even number, both of which make completing the square much easier. </li>
</ul>
<p>Some students may wonder why \(4a\) is chosen to complete the square when solving \(ax^2+bx+c=0\). Why not \(a\), \(9a\), or \(16a\)? Any of these would work. The equation could also have been divided by \(a\) to make the first term \(x^2\), which is a perfect square. (If time permits, consider asking students to try completing the square using one of these alternatives.) While these alternatives result in a perfect square for the first term, they may not give an even number for the coefficient of the second term or they may produce an equation with larger numbers, making it a bit trickier to manipulate.</p>
<ul class="os-raise-noindent">
<li> We can tell what constant term is needed to complete the square by dividing the coefficient of the linear term by the positive square root of the coefficient of \(x^2\), taking half of it, and squaring that number. In this case: </li>
<ul>
<li> The coefficient of \(x\) is \(4ab\). </li>
<li> The coefficient of \(x^2\) is \(4a^2\), so the square root is \(2a\). </li>
<li> Dividing \(4ab\) by \(2a\) gives \(2b\). Half of \(2b\) is \(b\). </li>
<li> The constant term that completes the square is \(b^2\). </li>
</ul>
</ul>
<p>Emphasize that the quadratic formula essentially captures the steps for completing the square in one expression. Every time we solve a quadratic equation by completing the square, we are essentially using the quadratic formula, but in a less condensed way.</p>
<h3>9.8.2: 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>Looking at the steps of deriving the quadratic formula, which step should follow the one below?</p>
<p>\((x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}\)</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>Complete the square.</p>
</td>
<td>
<p>Incorrect. Let's try again a different way: This was already done for the perfect square trinomial to be factored. The correct answer is to take the square root of both sides.</p>
</td>
</tr>
<tr>
<td>
<p>Take the square root of both sides.</p>
</td>
<td>
<p>That's correct! Check yourself: To get rid of the power of 2, take the square root of both sides and remember the plus and minus sign on the right side.</p>
</td>
</tr>
<tr>
<td>
<p>Multiply both sides by \(4a^2\).</p>
</td>
<td>
<p>Incorrect. Let's try again a different way: In a previous step, both sides were already divided by \(4a^2\). The correct answer is to take the square root of both sides.</p>
</td>
</tr>
<tr>
<td>
<p>Subtract \(\frac{b}{2a}\) from both sides.</p>
</td>
<td>
<p>Incorrect. Let's try again a different way: This cannot be done until after the square root is taken. The correct answer is to take the square root of both sides.</p>
</td>
</tr>
</tbody>
</table>
<br>
<!--END SELF CHECK INTRO BEFORE Tables -->
<h3>9.8.2: 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>A General Way to Derive the Quadratic Formula</h4>
<p>Here is another way to look at how to derive the quadratic formula:</p>
<p>We can derive the quadratic formula by completing the square. We will assume that the leading coefficient is positive; if it is negative, we can multiply the equation by −1 and obtain a positive \(a\). Given \(ax^2+bx+c=0\), \(a \neq 0\), we will complete the square as follows:</p>
<p><strong>Step 1</strong> - First, move the constant term to the right side of the equal sign:<br>
\(ax^2+bx= -c\)</p>
<p><strong>Step 2</strong> - As we want the leading coefficient to equal 1, divide through by \(a\):<br>
\(x^2+\frac{b}{a}x= -\frac{c}{a}\)</p>
<p><strong>Step 3</strong> - Then, find \(\frac{1}{2}\) of the middle term, and add \((\frac{1}{2} \cdot \frac{b}{a})^2=\frac{b^2}{4a^2}\) to both sides of the equal sign:<br>
\(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}−\frac{c}{a}\)</p>
<p><strong>Step 4</strong> - Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:<br>
\((x+\frac{b}{2a})^2=\frac{b^2−4ac}{4a^2}\)</p>
<p><strong>Step 5</strong> - Now, use the square root property, which gives:</p>
<p>\(x+\frac b{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}\)</p>
<p>\(x+\frac b{2a}=\frac{\pm\sqrt{b^2-4ac}}{2a}\)</p>
<p><strong>Step 6</strong> - Finally, add \(-\frac{b}{2a}\) to both sides of the equation and combine the terms on the right side. Thus:
</p>
<p> \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)</p>
<h4>Try It: A General Way to Derive the Quadratic Formula</h4>
<p>Using the general way to derive the quadratic formula, find the step after</p>
<p>\(x+\frac b{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}\).</p>
<p>Write down your answer, then select the <strong>solution</strong> button to check your work.</p>
<h5>Solution</h5>
<p>Here is how to find the next step:</p>
<p>When deriving the quadratic formula, the object is to get the \(x\) alone.</p>
<p>After \(x+\frac b{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}\), the \(\frac{b}{2a}\) must be subtracted from each side. Then \(x=-\frac b{2a}\pm\sqrt{\frac{b^2-4ac}{4a^2}}\) can be simplified.</p>