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<h4>Activity (25 minutes)</h4>
<p>In this activity, students encounter quadratic equations that are slightly more complex than those in the warm up.
</p>
<p>Encourage students to first isolate the squared variable algebraically. Then have them apply the square root
property, while reinforcing that this yields two solutions.</p>
<h4>Launch</h4>
<p>Introduce the square root property as shown at the top of the activity. Emphasize that when using the property to
solve an equation, there is always a positive and a negative square root.</p>
<p>Consider arranging students in groups of two and asking them to work quietly for a few minutes before discussing
their thinking 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>
<p class="os-raise-extrasupport-name">MLR 7 Compare and Connect: Representing, Conversing</p>
</div>
<div class="os-raise-extrasupport-body">
<p>Use this routine to prepare students for the
whole-class
discussion. At the appropriate time, invite student pairs to create a visual display of their strategies for solving
the quadratic equations. Allow students time to quietly circulate and analyze the strategies in at least two other
visual displays in the room. Give students quiet think time to consider what is the same and what is different about
their solution strategies. Next, ask students to return to their partner and discuss what they noticed. Listen for
and amplify observations that highlight advantages and disadvantages to each method. This will help students make
connections between different strategies for solving quadratic equations.</p>
<p class="os-raise-text-italicize">Design Principle(s): Optimize output; Cultivate conversation</p>
<p class="os-raise-extrasupport-title">Learn more about this routine</p>
<p>
<a href="https://www.youtube.com/watch?v=PF8fRA107OA;&rel=0" target="_blank">View the instructional video</a>
and
<a href="https://k12.openstax.org/contents/raise/resources/94a1159e7b81493c647515711f325771076d99b8" target="_blank">follow along with the materials</a>
to assist you with learning this routine.
</p>
<p class="os-raise-extrasupport-title">Provide support for students</p>
<p>
<a href="https://k12.openstax.org/contents/raise/resources/0b8a1a4ac3425e84a1d5452b3a5dffa38deb6b13" target="_blank">Distribute graphic organizers</a>
to the students to assist them with participating in this routine.
</p>
</div>
</div>
<br>
<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">Engagement: Develop Effort and Persistence</p>
</div>
<div class="os-raise-extrasupport-body">
<p>
Chunk this task into manageable parts for students who benefit
from
support with organizational skills in problem solving. For example, cut problems into four separate slips and have
students obtain the new problem only after finishing the last. Provide students four copies of a graphic organizer
with two sections—one for work, and another for notes/explanations. Invite students to record their work for each
problem on its own paper. When presenting multiple strategies and approaches, encourage students to use the space
for notes/explanations to record explanations and alternate strategies. </p>
<p class="os-raise-text-italicize">Supports accessibility for: Memory; Organization</p>
</div>
</div>
<br>
<h4>Student Activity</h4>
<p>A whole number that has two factors that are equal to each other is called a perfect square. Those equal factors that
multiply to be the number are called the <span class="os-raise-ib-tooltip" data-schema-version="1.0"
data-store="glossary-tooltip">square root of a number</span>. For instance, 3 is a square root of 9 because \(3 \cdot 3 =
9.\)</p>
<p>Let’s look at the square root property.</p>
<br>
<div class="os-raise-graybox">
<p class="os-raise-text-bold">SQUARE ROOT PROPERTY</p>
<p>If \(x^2 = k\), then</p>
<p class="os-raise-noindent">\(x=\sqrt{k} \quad \text { and } \quad x=-\sqrt{k} \rightarrow x=\pm \sqrt{k}\).</p>
</div>
<br>
<p>Each of these equations has two solutions. Use your knowledge of square roots and the square root property. What are
the solutions? Explain or show your reasoning.</p>
<ol class="os-raise-noindent">
<li>\(n^2 + 4=404\)</li>
</ol>
<p><strong>Answer:</strong> \(n=20\) and \(n=-20\)</p>
<p>If \(n^2 + 4 = 404\), then \(n^2 = 400\).<br>
Using the square root property, \(n= \pm \sqrt{400}\).<br>
Therefore, \(n=20\) and \(n=-20\).</p>
<ol class="os-raise-noindent" start="2">
<li>\(432=3n^2\)</li>
</ol>
<p><strong>Answer:</strong> \(n=12\) and \(n=-12\)</p>
<p>Divide each side of the equation by 3.<br>
\(n^2=144\)<br>
Using the square root property, \(n= \pm \sqrt{144}\).<br>
Therefore, \(n=12\) and \(n=-12\).</p>
<ol class="os-raise-noindent" start="3">
<li>\(\frac{n^2}{5}−20=0\)</li>
</ol>
<p><strong>Answer:</strong> \(n=10\) and \(n=-10\)</p>
<p>Add 20 to each side of the equation.<br>
Then multiply each side of the equation by 5.<br>
\(n^2 = 100\)<br>
Using the square root property, \(n= \pm \sqrt{100}\).<br>
Therefore, \(n=10\) and \(n=-10\).</p>
<ol class="os-raise-noindent" start="4">
<li>\(2n^2−7=65\)</li>
</ol>
<p><strong>Answer:</strong> \(n=6\) and \(n=-6\)</p>
<p>Add 7 to each side of the equation.<br>
Divide each side of the equation by 2.<br>
\(n^2=36\)<br>
Using the square root property, \(n= \pm \sqrt{36}\).<br>
Therefore, \(n=6\) and \(n=-6\).</p>
<h4>Student Facing Extension</h4>
<h5>Are you ready for more?</h5>
<ol class="os-raise-noindent">
<li>How many solutions does the equation \((x −3)(x+1)(x+5)=0\) have? What are the solutions?</li>
</ol>
<p><strong>Answer:</strong> Three solutions: \(x=3\), \(x=-1\), and \(x=-5\).</p>
<ol class="os-raise-noindent" start="2">
<li>How many solutions does the equation \((x −2)(x−7)(x−2)=0\) have? What are the solutions?</li>
</ol>
<p><strong>Answer:</strong> Two solutions: \(x=2\) and \(x=7\).</p>
<ol class="os-raise-noindent" start="3">
<li>Write a new equation that has 10 solutions.</li>
</ol>
<p><strong>Answer:</strong> There are many different possible answers. For example:</p>
<p>\((x −1)(x −2)(x −3)(x −4)(x −5)(x −6)(x −7)(x −8)(x −9)(x
−10)=0\).</p>
<h4>Activity Synthesis</h4>
<p>Ask students to explain in their own words why the square root property is true. Use an example such as \(x^2=64\) to
help explain. Ask a student to complete the following sentence:</p>
<p class="os-raise-noindent">If \(x^2=64\), then ______________________.</p>
<p>Sample answers:</p>
<ul class="os-raise-noindent">
<li> If \(x^2=64\), then \(x = \sqrt{64}\) and \(x =- \sqrt{64}\). </li>
<li> If \(x^2=64\), then \(x= \pm \sqrt{64}\). </li>
<li> If \(x^2=64\), then \(x=8\) and \(x=-8\). </li>
</ul>
<p>Emphasize that there are two solutions that make the equation true, a positive square root and a negative square
root.</p>
<h3>8.3.2: Self Check</h3>
<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>Solve for \(p\). Find all solutions.</p>
<p>\(3p^2+8 =251\)</p>
<table class="os-raise-textheavytable">
<thead>
<tr>
<th scope="col">Answers</th>
<th scope="col">Feedback</th>
</tr>
</thead>
<tbody>
<tr>
<td>
\(p=9\)
</td>
<td>
Incorrect. Let’s try again a different way: Subtract 8 from both sides. Divide both sides by 3. This
leaves \(p^2=81\). When you take the square root of 81, there is more than one solution. The answer is \(p=9\)
and \(p=-9\).
</td>
</tr>
<tr>
<td>
\(p=11\) and \(p=-11\)
</td>
<td>
Incorrect. Let’s try again a different way: First, subtract 8 from both sides of the equation. Then
divide both sides by 3. Take the square root of 81. The answer is \(p=9\) and \(p=-9\).
</td>
</tr>
<tr>
<td>
\(p=11\)
</td>
<td>
Incorrect. Let’s try again a different way: First, subtract 8 from both sides of the equation. Then
divide both sides by 3. Take the square root of 81. The answer is \(p=9\) and \(p=-9\).
</td>
</tr>
<tr>
<td>
<p>\(p=9\) and \(p=-9\)
</td>
<td>
That’s correct! Check yourself: Insert each value into the original equation. \(3(9)^2+8=251\) is true,
and \(3(-9)^2+8=251\) is true.
</td>
</tr>
</tbody>
</table>
<br>
<h3>8.3.2: Additional Resources</h3>
<p class="os-raise-text-bold"><em>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.</em></p>
<h4>Using the Square Root Property to Solve Quadratic Equations </h4>
<p>Let’s look again at the square root property.</p>
<br>
<div class="os-raise-graybox">
<p class="os-raise-text-bold">SQUARE ROOT PROPERTY</p>
<p>If \(x^2=k\), then</p>
<p class="os-raise-noindent">\(x=\sqrt{k} \quad \text { and } \quad x=-\sqrt{k} \rightarrow x=\pm \sqrt{k}\).</p>
</div>
<br>
<p>Notice that the square root property gives two solutions to an equation of the form \(x^2=k\), the principal square
root of \(k\) and its opposite. We could also write the solution as \(x= \pm \sqrt{k}\). We read this as \(x\) equals
positive and negative the square root of \(k\).</p>
<p class="os-raise-text-bold">Example 1</p>
<p>Let’s solve \(x^2= 9\) using the square root property.</p>
<p>\(x^2=9\)</p>
<p>\(x= \pm \sqrt{9}\)</p>
<p>\(x= \pm 3\)</p>
<p>So, \(x=3\) and \(x=-3\).</p>
<p>If the number in the radical is not a perfect square, leave that part inside the radical. Let’s solve another
example.</p>
<p class="os-raise-text-bold">Example 2</p>
<p>Solve \(x^2−50=0\).</p>
<p><strong>Step 1</strong> - Isolate the quadratic term and make its coefficient 1. </p>
<p>To do this, add 50 to both sides of the equation to get \(x^2\)
by itself.</p>
<p>\(x^2−50=0\) </p>
<p>\(x^2=50\)</p>
<p><strong>Step 2</strong> - Use the square root property. </p>
<p>Remember to write the symbol.</p>
<p>\(x = \pm \sqrt{50}\) </p>
<p><strong>Step 3</strong> - Simplify the radical by pulling out the factors that are perfect squares.</p>
<p>\(x = \pm \sqrt{25} \cdot \sqrt{2}\)</p>
<p>\(x = \pm 5 \sqrt{2}\)</p>
<p>\(x =5 \sqrt{2}\), \(x =−5 \sqrt{2}\)</p>
<p><strong>Step 4</strong> - Check the solutions.</p>
<p>\(\begin{aligned}
x^{2}-50=0 \\
({\style{color:red}5}
\sqrt{{\style{color:red}2}})^{2}-50\stackrel{?}{=}0 \\
25 \cdot 2-50\stackrel{?}{=}0 \\
0=0 {\style{color:red}\checkmark} \\
x^{2}-50=0 \\
({\style{color:red}-}{\style{color:red}5} \sqrt{{\style{color:red}2}})^{2}-50\stackrel{?}{=}0 \\
25 \cdot 2-50\stackrel{?}{=}0 \\
0=0 {\style{color:red}\checkmark} \\
\end{aligned}\)</p>
<p>The steps for solving a quadratic equation using the square root property are listed here.</p>
<br>
<div class="os-raise-graybox">
<p class="os-raise-text-bold">HOW TO…</p>
<p>Solve a quadratic equation using the square root property.</p>
<p><strong>Step 1</strong> - Isolate the quadratic term and make its coefficient 1.</p>
<p><strong>Step 2</strong> - Use the square root property.</p>
<p><strong>Step 3</strong> - Simplify the radical.</p>
<p><strong>Step 4 </strong>
Check the solutions.</p>
</div>
<br>
<p>Before using the square root property in Step 2, the coefficient of the variable term must equal 1. In the next
example, we must divide both sides of the equation by the coefficient 3 before using the square root property.</p>
<p class="os-raise-text-bold">Example 3</p>
<p>Solve \(3z^2=108\).</p>
<p><strong>Step 1</strong> - Isolate the quadratic term and make its coefficient 1. </p>
<p>When the quadratic term is isolated, divide by 3 to make its
coefficient 1. Simplify.</p>
<p>\(3z^2=108\)</p>
<p>\(\frac{3z^2}{3}=\frac{108}{3}\) </p>
<p> \(z^2=36\) </p>
<p><strong>Step 2</strong> - Use the square root property.</p>
<p> \(z = \pm \sqrt{36}\) </p>
<p><strong>Step 3</strong> - Simplify the radical. </p>
<p>Rewrite to show two solutions.</p>
<p> \(z= \pm 6\) </p>
<p>\(z=6\), \(z=-6\) </p>
<p><strong>Step 4</strong> - Check the solutions.</p>
<p>\(\begin{array}{rr} 3 z^{2}=108 & 3 z^{2}=108 \\ 3({\style{color:red}6})^{2} \stackrel{?}{=} 108 &
3({\style{color:red}-}{\style{color:red}6})^{2} \stackrel{?}{=} 108 \\ 3(36) \stackrel{?}{=} 108 & 3(36)
\stackrel{?}{=} 108 \\ 108=108 {\style{color:red}\checkmark} & 108=108 {\style{color:red}\checkmark} \end{array}
\)</p>
<h4>Try It: Using the Square Root Property to Solve Quadratic Equations</h4>
<p>Solve the following quadratic equations using the square root property.</p>
<ol class="os-raise-noindent">
<li>\(2n^2=98\) </li>
</ol>
<ol class="os-raise-noindent" start="2">
<li> \(\frac{n^2}{2}−30=68\) </li>
</ol>
<p>Write down your answers, then select the <strong>solution</strong> button to compare your work.</p>
<h5>Solution</h5>
<p>Here is how to solve these quadratic equations using the square root property:</p>
<ol class="os-raise-noindent">
<li> \(\begin{aligned} 2 n^{2} &=98 \\ \frac{2 n^{2}}{2} &=\frac{98}{2} \\ n^{2} &=49 \\ n &=\pm
\sqrt{49} \\ n &=7 \text { and } n=-7 \end{aligned} \)
</li>
</ol> <br>
<ol class="os-raise-noindent" start="2">
<li> \(\begin{aligned} \frac{n^{2}}{2}-30 &=68 \\ \frac{n^{2}}{2} &=98 \\ n^{2} &=196 \\ n &=\pm
\sqrt{196} \\ n=14 \text { and } n &=-14 \end{aligned} \)</li>
</ol>