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<title>Real World Examples of Quadratic Equations</title>
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<div id="content" role="main"><!-- #BeginEditable "Body" -->
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<h1 class="center">Real World Examples of<br>
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Quadratic Equations</h1>
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<p>A <b>Quadratic Equation</b> looks like this:</p>
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<p align="center"><img src="images/quadratic-equation-reason.svg" alt="Quadratic Equation"></p>
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<p><a href="quadratic-equation.html">Quadratic equations</a> pop up in many real world situations! </p>
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<p>Here we have collected some examples for you, and solve each using different methods: </p>
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<ul>
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<li><a href="factoring-quadratics.html">Factoring Quadratics</a></li>
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<li><a href="completing-square.html">Completing the Square</a></li>
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<li><a href="quadratic-equation-graphing.html">Graphing Quadratic Equations</a></li>
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<li><a href="quadratic-equation.html">The Quadratic Formula</a></li>
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<li><a href="../quadratic-equation-solver.html">Online Quadratic Equation Solver</a></li>
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</ul>
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<p>Each example follows three general stages:</p>
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<ul>
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<li>Take the real world description and make some equations </li>
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<li>Solve!</li>
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<li>Use your common sense to interpret the results</li>
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</ul>
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<p> </p>
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<p style="float:left; margin: 0 10px 5px 0;"><img src="images/ball-throw.jpg" alt="ball throw" height="283" width="110"></p>
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<h2>Balls, Arrows, Missiles and Stones</h2>
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<p>When you throw a ball (or shoot an arrow, fire a missile or throw a stone) it goes up into the air, slowing as it travels, then comes down again faster and faster ...</p>
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<p class="center larger">... and a <a href="quadratic-equation.html">Quadratic Equation</a> tells you its position at all times!</p>
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<p> </p>
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<h2>Example: Throwing a Ball</h2>
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<h3>A ball is thrown straight up, from 3 m above the ground, with a velocity of 14 m/s. When does it hit the ground?</h3>
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<div class="simple">
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<p>Ignoring air resistance, we can work out its height by adding up these three things:<br>
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(Note: <b>t</b> is time in seconds)</p>
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<table align="center" border="0">
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<tbody>
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<tr>
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<td align="right">The height starts at 3 m:</td>
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<td width="20"> </td>
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<td class="larger" align="center">3</td>
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</tr>
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<tr>
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<td align="right">It travels upwards at 14 meters per second (14 m/s):</td>
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<td> </td>
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<td class="larger" align="center">14t</td>
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</tr>
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<tr>
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<td align="right">Gravity pulls it down, changing its position by <i>about</i> 5 m per second squared:</td>
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<td> </td>
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<td class="larger" align="center" valign="bottom">−5t<sup>2</sup></td>
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</tr>
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<tr>
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<td><i>(Note for the enthusiastic: the <span class="larger">-5t<sup>2</sup></span> is simplified from <span class="larger">-(½)at<sup>2</sup></span> with a=9.8 m/s<sup>2</sup>)</i></td>
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<td> </td>
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<td class="larger" align="center" valign="bottom"> </td>
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</tr>
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</tbody></table>
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</div>
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<p>Add them up and the height <b>h</b> at any time <b>t</b> is: </p>
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<p class="center larger">h = 3 + 14t − 5t<sup>2</sup></p>
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<p> And the ball will hit the ground when the height is zero:</p>
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<p class="center larger"> 3 + 14t − 5t<sup>2</sup> = 0</p>
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<p>Which is a <a href="factoring-quadratics.html">Quadratic Equation</a> ! </p>
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<p>In "Standard Form" it looks like:</p>
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<p class="center larger">−5t<sup>2</sup> + 14t + 3 = 0</p>
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<p>It looks even better when we <span class="left">multiply all terms by −1</span>:</p>
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<p class="center larger">5t<sup>2</sup> − 14t − 3 = 0</p>
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<p>Let us solve it ...</p>
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<p> </p>
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<p>There are many ways to solve it, here we will factor it using the "Find two numbers that
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multiply to give <b>a×c</b>, and add to give <b>b</b>" method in <a href="factoring-quadratics.html">Factoring Quadratics</a>:</p>
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<p class="center">a×c = <b><span class="larger">−</span>15</b>, and b = <b><span class="larger">−</span>14</b>. </p>
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<p>The factors of −15 are: −15, −5, −3, −1, 1, 3, 5, 15</p>
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<p>By trying a few combinations we find that <b>−15</b> and <b>1</b> work
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(−15×1 = −15,
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and −15+1 = −14)</p>
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<div class="tbl">
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<div class="row"><span class="left">Rewrite middle with −15 and 1:</span><span class="right">5t<sup>2</sup> <span class="hilite">− 15t + t</span> − 3 = 0</span></div>
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<div class="row"><span class="left">Factor first two and last two:</span><span class="right">5t(t − 3) + 1(t − 3) = 0</span></div>
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<div class="row"><span class="left">Common Factor is (t − 3):</span><span class="right">(5t + 1)(t − 3) = 0</span></div>
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<div class="row"><span class="left">And the two solutions are:</span><span class="right">5t + 1 = 0 or t − 3 = 0</span></div>
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<div class="row"><span class="left"> </span><span class="right">t = <b>−0.2</b> or t = <b>3</b></span></div>
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</div>
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<p>The "t = −0.2" is a negative time, impossible in our case.</p>
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<p>The "t = 3" is the answer we want:</p>
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<p class="center larger">The ball hits the ground after 3 seconds!</p>
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<p style="float:left; margin: 0 10px 5px 0;"><img src="images/quadratic-1a.gif" alt="quadratic graph ball" height="309" width="174"></p>
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<p>Here is the graph of the <a href="../geometry/parabola.html">Parabola</a><span class="larger"> h = −5t<sup>2</sup> + 14t + 3</span></p>
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<p>It shows you the <b>height</b> of the ball vs <b>time</b></p>
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<p>Some interesting points:</p>
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<p><span class="large">(0,3)</span> When t=0 (at the start) the ball is at 3 m</p>
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<p><span class="large">(−0.2,0)</span> says that −0.2 seconds BEFORE we threw the ball it was at ground level. This never happened! So our common sense says to ignore it.</p>
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<p> <span class="large">(3,0)</span> says that at 3 seconds the ball is at ground level.</p>
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<p>Also notice that the ball goes <b>nearly 13 meters</b> high.</p>
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<div style="clear:both"></div>
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<div class="center80">
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<p>Note: You can find exactly where the top point is! </p>
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<p>The method is explained in <a href="quadratic-equation-graphing.html">Graphing Quadratic Equations</a>, and has two steps:</p>
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<p>Find where (along the horizontal axis) the top occurs using <b>−b/2a</b>:</p>
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<ul>
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<li>t = −b/2a = −(−14)/(2 × 5) = 14/10 = <b>1.4 seconds</b></li>
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</ul>
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<p>Then find the height using that value (1.4)</p>
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<ul>
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<li>h = −5t<sup>2</sup> + 14t + 3 = −5(1.4)<sup>2</sup> + 14 × 1.4 + 3 = <b>12.8 meters</b></li>
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</ul>
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<p>So the ball reaches the highest point of 12.8 meters after 1.4 seconds.</p>
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</div><p> </p>
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<table border="0">
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<tbody>
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<tr>
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<td><img src="images/bike.jpg" alt="bike" height="100" width="154"></td>
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<td> </td>
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<td>
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<h2>Example: New Sports Bike</h2>
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<p>You have designed a new style of sports bicycle!</p>
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<p>Now you want to make lots of them and sell them for profit.</p></td>
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</tr>
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</tbody></table>
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<p>Your <b>costs</b> are going to be:</p>
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<ul>
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<li>$700,000 for manufacturing set-up costs, advertising, etc</li>
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<li>$110 to make each bike</li>
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</ul>
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<p style="float:right; margin: 0 0 5px 10px;"><img src="images/graph-bike-demand.svg" alt="graph bike demand curve"></p>
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<p>Based on similar bikes, you can expect <b>sales</b> to follow this "Demand Curve":</p>
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<ul>
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<li>Unit Sales = 70,000 − 200P</li>
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</ul>
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<p>Where "P" is the price. </p>
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<p>For example, if you set the price:</p>
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<ul>
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<li>at $0, you just give away 70,000 bikes </li>
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<li>at $350, you won't sell any bikes at all</li>
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<li>at $300 you might sell <b>70,000 − 200×300 = 10,000</b> bikes</li>
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</ul>
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<p>So ... what is the best price? And how many should you make? </p>
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<p><b>Let us make some equations!</b></p>
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<p>How many you sell depends on price, so use "P" for Price as the variable</p>
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<ul>
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<li>Unit Sales = 70,000 − 200P</li>
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<li>Sales in Dollars = Units × Price = (70,000 − 200P) × P = 70,000P − 200P<sup>2</sup></li>
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<li>Costs = 700,000 + 110 x (70,000 − 200P) = 700,000 + 7,700,000 − 22,000P = 8,400,000 − 22,000P</li>
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<li>Profit = Sales-Costs = 70,000P − 200P<sup>2</sup> − (8,400,000 − 22,000P) = −200P<sup>2</sup> + 92,000P − 8,400,000</li>
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</ul>
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<p class="large center">Profit = −200P<sup>2</sup> + 92,000P − 8,400,000</p>
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<p>Yes, a Quadratic Equation. Let us solve this one by <a href="completing-square.html">Completing the Square</a>.</p>
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<div class="example">
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<h3> Solve: −200P<sup>2</sup> + 92,000P − 8,400,000 = 0</h3>
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<p><b>Step 1</b> Divide all terms by -200</p>
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<div class="so"> P<sup>2</sup> – 460P + 42000 = 0</div>
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<p> <b>Step 2</b> Move the number term to the right side of the equation:</p>
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<div class="so">P<sup>2</sup> – 460P = -42000</div>
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<p><b>Step 3</b> Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation:</p>
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<p>(b/2)<sup>2</sup> = (−460/2)<sup>2</sup> = (−230)<sup>2</sup> = 52900</p>
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<div class="so">P<sup>2</sup> – 460P + 52900 = −42000 + 52900</div>
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<div class="so">(P – 230)<sup>2</sup> = 10900</div>
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<p><b>Step 4</b> Take the square root on both sides of the equation:</p>
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<div class="so"> P – 230 = ±√10900 = ±104 (to nearest whole number)</div>
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<p><b>Step 5</b> Subtract (-230) from both sides (in other words, add 230):</p>
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<div class="so"> P = 230 ± 104 = 126 or 334</div>
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</div>
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<p>What does that tell us? It says that the profit is ZERO when the Price is $126 or $334</p>
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<p>But we want to know the maximum profit, don't we? </p>
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<p><b>It is exactly half way in-between!</b><span class="larger"> At $230</span></p>
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<p>And here is the graph:</p>
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<p class="center larger"><img src="images/graph-bike-profit.svg" alt="graph bike profit best"><br>
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Profit = −200P<sup>2</sup> + 92,000P − 8,400,000</p>
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<p>The best sale price is <b>$230</b>, and you can expect:</p>
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<ul>
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<li>Unit Sales = 70,000 − 200 x 230 = 24,000 </li>
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<li>Sales in Dollars = $230 x 24,000 = $5,520,000</li>
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<li>Costs = 700,000 + $110 x 24,000 = $3,340,000</li>
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<li>Profit = $5,520,000 − $3,340,000 = <b>$2,180,000</b></li>
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</ul>
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<p>A very profitable venture.</p>
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<h2>Example: Small Steel Frame</h2>
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<p style="float:left; margin: 0 10px 5px 0;"><img src="images/quadratic-4a.gif" alt="area=28" height="121" width="171"></p>
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<p>Your company is going to make frames as part of a new product they are launching.</p>
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<p>The frame will be cut out of a piece of steel, and to keep the weight down, the final area should be <b>28 cm<sup>2</sup></b></p>
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<p>The inside of the frame has to be<b> 11 cm by 6 cm</b></p>
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<p>What should the width <b>x</b> of the metal be?</p>
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<p>Area of steel before cutting:</p>
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<div class="so"> Area = (11 + 2x) × (6 + 2x) cm<sup>2</sup></div>
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<div class="so"> Area = 66 + 22x + 12x + 4x<sup>2</sup></div>
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<div class="so"> Area = 4x<sup>2</sup> + 34x + 66 </div>
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<p>Area of steel after cutting out the 11 × 6 middle:</p>
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<div class="so"> Area = 4x<sup>2</sup> + 34x + 66 − 66</div>
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<div class="so"> Area = 4x<sup>2</sup> + 34x</div>
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<p style="float:right; margin: 0 0 5px 10px;"><img src="images/quadratic-4c.svg" alt="quadratic 4x^2 + 34x"></p>
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<h3>Let us solve this one <a href="quadratic-equation-graphing.html">graphically</a>!</h3>
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<p class="right">Here is the graph of <span class="larger">4x<sup>2</sup> + 34x</span> :</p>
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<p>The desired area of <b>28</b> is shown as a horizontal line.</p>
|
||
<p>The area equals 28 cm<sup>2</sup> when:</p>
|
||
<p class="center"><b>x is <i>about</i> −9.3 or 0.8</b></p>
|
||
<p>The negative value of <b>x</b> make no sense, so the answer is:</p>
|
||
<p class="large center">x = 0.8 cm (approx.)</p>
|
||
<p> </p>
|
||
|
||
<h2>Example: River Cruise</h2>
|
||
<h3>A 3 hour river cruise goes 15 km upstream and then back again. The river has a current of 2 km an hour. What is the boat's speed and how long was the upstream journey? </h3>
|
||
<p style="float:left; margin: 0 30px 5px 0;"><img src="images/river-boat.svg" alt="river sketch"></p>
|
||
<p>There are two speeds to think about: the speed the boat makes in the water, and the speed relative to the land:</p>
|
||
<ul>
|
||
<li>Let <b>x</b> = the boat's speed in the water (km/h)</li>
|
||
<li>Let <b>v</b> = the speed relative to the land (km/h)</li>
|
||
</ul>
|
||
<p>Because the river flows downstream at 2 km/h:</p>
|
||
<ul>
|
||
<li>when going upstream, <b>v = x−2</b> (its speed is reduced by 2 km/h)</li>
|
||
<li>when going downstream, <b>v = x+2</b> (its speed is increased by 2 km/h)</li>
|
||
</ul>
|
||
<p>We can turn those speeds into times using:</p>
|
||
<p class="larger" align="center">time = distance / speed</p>
|
||
<p align="center">(to travel 8 km at 4 km/h takes 8/4 = 2 hours, right?)</p>
|
||
<p>And we know the total time is 3 hours:</p>
|
||
<p class="larger" align="center">total time = time upstream + time downstream = 3 hours</p>
|
||
<p>Put all that together:</p>
|
||
<p align="center"><span class="larger">total time = 15/(x−2) + 15/(x+2) = 3 hours</span></p>
|
||
<p>Now we use our algebra skills to solve for "x".</p>
|
||
<p>First, get rid of the fractions by multiplying through by <b>(x-2)</b><b>(x+2)</b>: </p>
|
||
<p class="larger" align="center">3(x-2)(x+2) = 15(x+2) + 15(x-2)</p>
|
||
<p>Expand everything:</p>
|
||
<p class="larger" align="center">3(x<sup>2</sup>−4) = 15x+30 + 15x−30</p>
|
||
<p>Bring everything to the left and simplify:<b></b></p>
|
||
<p class="larger" align="center">3x<sup>2</sup> − 30x − 12 = 0</p>
|
||
<p>It is a Quadratic Equation! Let us solve it using the <a href="quadratic-equation.html">Quadratic Formula</a>:</p>
|
||
<p class="center"><img src="images/quadratic-formula.svg" alt="Quadratic Formula: x = [ -b (+-) sqrt(b^2 - 4ac) ] / 2a"></p>
|
||
<p class="center">Where <b>a</b>, <b>b</b> and <b>c</b> are
|
||
from the <br>
|
||
Quadratic Equation in "Standard Form": <b>ax<sup>2</sup> + bx + c = 0</b></p>
|
||
<div class="example">
|
||
<h3>Solve 3x<sup>2</sup> - 30x - 12 = 0</h3>
|
||
<div class="tbl">
|
||
<div class="row"><span class="left"><b>Coefficients are:</b></span><span class="right"><span class="center"><b>a = 3</b>, <b>b = −30</b> and <b>c = −12</b></span></span></div>
|
||
<div class="row"><span class="left"><b>Quadratic Formula:</b></span><span class="right">x = [ −b ± √(b<sup>2</sup>−4ac) ] / 2a</span></div>
|
||
<div class="row"><span class="left"><b>Put in a, b and c:</b></span><span class="right">x = [ −(−30) ± √((−30)<sup>2</sup>−4×3×(−12)) ] / (2×3)</span></div>
|
||
<div class="row"><span class="left"><b>Solve</b>:</span><span class="right">x = [ 30 ± √(900+144) ] / 6</span></div>
|
||
<div class="row"><span class="left"> </span><span class="right">x = [ 30 ± √(1044) ] / 6</span></div>
|
||
<div class="row"><span class="left"> </span><span class="right">x = ( 30 ± 32.31 ) / 6 </span></div>
|
||
<div class="row"><span class="left"> </span><span class="right">x = −0.39 <b>or</b> 10.39</span></div>
|
||
</div>
|
||
<p> </p>
|
||
<p class="center "><span class="larger"><b>Answer:</b> x = −0.39 <b>or</b> 10.39</span> (to 2 decimal places)</p>
|
||
|
||
</div>
|
||
<p>x = −0.39 makes no sense for this real world question, but x = 10.39 is just perfect!</p>
|
||
<p align="center">Answer: <span class="large">Boat's Speed = 10.39 km/h </span>(to 2 decimal places)</p>
|
||
<p align="center">And so the upstream journey = 15 / (10.39−2) = 1.79 hours = <span class="large">1 hour 47min</span></p>
|
||
<p align="center">And the downstream journey = 15 / (10.39+2) = 1.21 hours = <span class="large">1 hour 13min</span></p>
|
||
<p> </p>
|
||
|
||
<h2>Example: Resistors In Parallel </h2>
|
||
<p>Two resistors are in parallel, like in this diagram:</p>
|
||
<p class="center"><img src="images/quadratic-3b.gif" alt="quadratic resistors R1 and R1+3" height="113" width="348"></p>
|
||
<p>The total resistance has been measured at 2 Ohms, and one of the resistors is known to be 3 ohms more than the other. </p>
|
||
<p>What are the values of the two resistors?</p>
|
||
<p>The formula to work out total resistance "R<sub>T</sub>" is: </p>
|
||
|
||
<p class="center larger"><span class="intbl"><em>1</em><strong>R<sub>T</sub></strong>
|
||
</span> = <span class="intbl">
|
||
<em>1</em><strong>R<sub>1</sub></strong></span> + <span class="intbl"><em>1</em><strong>R<sub>2</sub></strong></span></p>
|
||
<p>In this case, we have R<sub>T</sub> = 2 and R<sub>2</sub> = R<sub>1</sub> + 3 </p>
|
||
<p class="center larger"><span class="intbl">
|
||
<em>1</em>
|
||
<strong>2</strong>
|
||
</span> = <span class="intbl">
|
||
<em>1</em>
|
||
<strong>R<sub>1</sub></strong>
|
||
</span> + <span class="intbl">
|
||
<em>1</em>
|
||
<strong>R<sub>1</sub>+3</strong>
|
||
</span></p>
|
||
<p>To get <span class="left">rid of the fractions we
|
||
can multiply all terms by 2R<sub>1</sub>(R<sub>1</sub> + 3)</span> and then simplify:</p>
|
||
<div class="tbl">
|
||
<div class="row"><span class="left">Multiply all terms by 2R<sub>1</sub>(R<sub>1</sub> + 3):</span><span class="right"><span class="intbl"><em>2R<sub>1</sub>(R<sub>1</sub>+3)</em><strong>2</strong></span> = <span class="intbl"><em>2R<sub>1</sub>(R<sub>1</sub>+3)</em><strong>R<sub>1</sub></strong></span> + <span class="intbl"><em>2R<sub>1</sub>(R<sub>1</sub>+3)</em><strong>R<sub>1</sub>+3</strong></span></span></div>
|
||
<div class="row"><span class="left">Then simplify:</span><span class="right">R<sub>1</sub>(R<sub>1</sub> + 3) = 2(R<sub>1</sub> + 3) + 2R<sub>1</sub></span></div>
|
||
<div class="row"><span class="left">Expand: </span><span class="right">R<sub>1</sub><sup>2</sup> + 3R<sub>1</sub> = 2R<sub>1</sub> + 6 + 2R<sub>1</sub></span></div>
|
||
<div class="row"><span class="left">Bring all terms to the left:</span><span class="right">R<sub>1</sub><sup>2</sup> + 3R<sub>1</sub> − 2R<sub>1</sub> − 6 − 2R<sub>1</sub> = 0</span></div>
|
||
<div class="row"><span class="left">Simplify:</span><span class="right"><span class="center">R<sub>1</sub><sup>2</sup> − R<sub>1</sub> − 6 = 0</span></span></div>
|
||
</div>
|
||
|
||
|
||
<p>Yes! A Quadratic Equation !</p>
|
||
<p>Let us solve it using our <a href="../quadratic-equation-solver.html">Quadratic Equation Solver</a>. </p>
|
||
<ul>
|
||
<li>Enter 1, −1 and −6 </li>
|
||
<li>And you should get the answers −2 and 3</li>
|
||
</ul>
|
||
<p> <span class="center">R<sub>1</sub></span> cannot be negative, so <span class="center"><b>R<sub>1</sub></b></span><b> = 3 Ohms</b> is the answer.</p>
|
||
<p class="center larger"> The two resistors are 3 ohms and 6 ohms.</p>
|
||
|
||
<p> </p>
|
||
<h2>Others</h2>
|
||
<p> Quadratic Equations are useful in many other areas:</p>
|
||
<p style="float:left; margin: 0 10px 5px 0;"><img src="../geometry/images/parabolic-dish.jpg" alt="parabolic dish" height="109" width="116"></p>
|
||
<p>For a parabolic mirror, a reflecting telescope or a satellite dish, the shape is defined by a quadratic equation.</p>
|
||
<p>Quadratic equations are also needed when studying lenses and curved mirrors.</p>
|
||
<p>And many questions involving time, distance and speed need quadratic equations.</p>
|
||
<p> </p>
|
||
<div class="related">
|
||
<a href="quadratic-equation.html">Quadratic Equations</a>
|
||
<a href="factoring-quadratics.html">Factoring Quadratics</a>
|
||
<a href="completing-square.html">Completing the Square</a>
|
||
<a href="quadratic-equation-graphing.html">Graphing Quadratic Equations</a>
|
||
<a href="../quadratic-equation-solver.html">Quadratic Equation Solver</a>
|
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<a href="index.html">Algebra Index</a>
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