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317 lines
13 KiB
HTML
317 lines
13 KiB
HTML
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<!-- #BeginEditable "Body" -->
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<h1 class="center">Systems of Linear and Quadratic Equations</h1>
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<p> </p>
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<div class="simple">
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<table style="border: 0; margin:auto;">
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<tbody>
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<tr>
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<td bgcolor="#eef"><img src="images/linear-quadratic-a.svg" alt="linear " height="205" width="198"></td>
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<td><span class="center">A <a href="linear-equations.html">Linear Equation</a> is an <b>equation</b> of a <b>line</b>.</span></td>
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</tr>
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<tr>
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<td bgcolor="#eef"><img src="images/linear-quadratic-b.svg" alt="quadratic" height="205" width="198"></td>
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<td>A <a href="quadratic-equation.html">Quadratic Equation</a> is the equation of a <a href="../geometry/parabola.html">parabola</a><br>
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and has at least one variable squared
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(such as x<sup>2</sup>)<br>
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</td>
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</tr>
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<tr>
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<td bgcolor="#eef"><img src="images/linear-quadratic-c.svg" alt="linear and quadratic" height="205" width="198"></td>
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<td>And together they form a <b>System</b><br>
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of a Linear and a Quadratic Equation</td>
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</tr>
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</tbody></table>
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</div>
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<p> </p>
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<p>A <b>System</b> of those two equations can be solved (find where they intersect), either:</p>
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<ul>
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<li><b>Graphically</b> (by plotting them both on the <a href="../data/function-grapher.html">Function Grapher</a> and zooming in)</li>
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<li>or using <b>Algebra</b></li>
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</ul>
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<h2>How to Solve using Algebra</h2>
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<ul>
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<li>Make both equations into "y =" format</li>
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<li>Set them equal to each other</li>
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<li>Simplify into "= 0" format (like a standard Quadratic Equation)</li>
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<li>Solve the Quadratic Equation!</li>
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<li>Use the linear equation to calculate matching "y" values, so we get (x,y) points as answers</li>
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</ul>
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<p>An example will help:</p>
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<div class="example">
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<h3>Example: Solve these two equations:</h3>
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<ul>
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<li>y = x<sup>2</sup> - 5x + 7</li>
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<li>y = 2x + 1</li>
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</ul>
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<p> </p>
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<p class="larger">Make both equations into "y=" format:</p>
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<p>They are both in "y=" format, so go straight to next step</p>
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<p> </p>
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<p class="larger">Set them equal to each other</p>
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<div class="so">x<sup>2</sup> - 5x + 7 = 2x + 1 </div>
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<p> </p>
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<p class="larger">Simplify into "= 0" format (like a standard Quadratic Equation)</p>
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<div class="so">Subtract 2x from both sides: x<sup>2</sup> - 7x + 7 = 1</div>
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<div class="so">Subtract 1 from both sides: x<sup>2</sup> - 7x + 6 = 0 </div>
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<p> </p>
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<p class="larger">Solve the Quadratic Equation!</p>
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<p>(The hardest part for me)</p>
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<p>You can read how to <a href="quadratic-equation.html">solve Quadratic Equations</a>, but here we will <a href="factoring-quadratics.html">factor the Quadratic Equation</a>:</p>
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<div class="so">Start with: <b>x<sup>2</sup> - 7x + 6 = 0</b></div>
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<div class="so">Rewrite -7x as -x-6x: <b>x<sup>2</sup> - x - 6x + 6 = 0</b></div>
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<div class="so">Then: <b>x(x-1) - 6(x-1) = 0</b></div>
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<div class="so">Then: <b>(x-1)(x-6) = 0</b></div>
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<p style="float:right; margin: 10px;"><img src="images/linear-quadratic-1.gif" alt="linear and quadratic" height="226" width="120"></p>
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<p>Which gives us the solutions <b>x=1</b> and <b>x=6</b></p>
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<p> </p>
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<p> </p>
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<p class="larger">Use the linear equation to calculate matching "y" values, so we get (x,y) points as answers</p>
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<p>The matching y values are (also see Graph):</p>
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<ul>
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<li>for x=<b>1</b>: y = 2x+1 = <b>3</b></li>
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<li>for x=<b>6</b>: y = 2x+1 = <b>13</b></li>
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</ul>
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<p> </p>
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<p class="center">Our solution: the two points are <b>(1,3)</b> and <b>(6,13)</b></p>
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<div class="example2"></div>
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</div>
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<p>I think of it as three stages:</p>
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<div class="center80">
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<p class="center">Combine into Quadratic Equation ⇒ Solve the Quadratic ⇒ Calculate the points</p>
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</div>
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<h2>Solutions</h2>
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<p>There are three possible cases:</p>
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<ul>
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<li><b>No</b> real solution (happens when they never intersect)</li>
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<li><b>One</b> real solution (when the straight line just touches the quadratic)</li>
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<li><b>Two</b> real solutions (like the example above)</li>
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</ul>
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<p class="center"><img src="images/linear-quadratic-2.svg" alt="linear and quadratic different intersections" height="228" width="563"></p>
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<p>Time for another example!</p>
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<div class="example">
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<h3>Example: Solve these two equations:</h3>
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<ul>
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<li>y - x<sup>2</sup> = 7 - 5x</li>
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<li>4y - 8x = -21</li>
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</ul>
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<p> </p>
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<p class="larger">Make both equations into "y=" format:</p>
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<p>First equation is: y - x<sup>2</sup> = 7 - 5x</p>
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<div class="so">Add x<sup>2</sup> to both sides: <b>y = x<sup>2</sup> + 7 - 5x</b></div>
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<p>Second equation is: 4y - 8x = -21</p>
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<div class="so">Add 8x to both sides: 4y = 8x - 21</div>
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<div class="so">Divide all by 4: <b>y = 2x - 5.25</b></div>
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<p> </p>
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<p class="larger">Set them equal to each other</p>
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<div class="so">x<sup>2</sup> - 5x + 7 = 2x - 5.25</div>
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<p> </p>
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<p class="larger">Simplify into "= 0" format (like a standard Quadratic Equation)</p>
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<div class="so">Subtract 2x from both sides: x<sup>2</sup> - 7x + 7 = -5.25</div>
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<div class="so">Add 5.25 to both sides: x<sup>2</sup> - 7x + 12.25 = 0 </div>
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<p> </p>
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<p class="larger">Solve the Quadratic Equation!</p>
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<p>Using the Quadratic Formula from <a href="quadratic-equation.html"> Quadratic Equations</a>:</p>
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<p style="float:right; margin: 10px;"> </p>
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<p style="float:right; margin: 10px;"><img src="images/linear-quadratic-3.gif" alt="linear and quadratic one intersection" height="201" width="145"></p>
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<ul>
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<li>x = [ -b ± √(b<sup>2</sup>-4ac) ] / 2a</li>
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<li>x = [ 7 ± √((-7)<sup>2</sup>-4×1×12.25) ] / 2×1</li>
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<li>x = [ 7 ± √(49-49) ] / 2</li>
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<li>x = [ 7 ± √0 ] / 2</li>
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<li><b>x = 3.5</b></li>
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</ul>
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<p>Just one solution! (The "discriminant" is 0)</p>
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<p> </p>
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<p class="larger">Use the linear equation to calculate matching "y" values, so we get (x,y) points as answers</p>
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<p>The matching y value is:</p>
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<ul>
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<li>for x=<b>3.5</b>: y = 2x-5.25 = <b>1.75</b></li>
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</ul>
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<p> </p>
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<p class="center">Our solution:<b> (3.5,1.75)</b></p>
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<div class="example2"></div>
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</div>
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<p> </p>
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<div class="example">
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<img style="float:right; margin: 0 0 5px 10px;" src="images/cannon.jpg" alt="Cannon">
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<h3>Real World Example</h3>
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<p class="larger">Kaboom!</p>
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<p>The cannon ball flies through the air, following a parabola: </p>
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<p><span class="larger">y = 2 + 0.12x - 0.002x<sup>2</sup></span></p>
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<p>The land slopes upward: <span class="larger">y = 0.15x </span></p>
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<p>Where does the cannon ball land?</p>
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<p class="center"><img src="images/linear-quadratic-5.gif" alt="linear quadratic cannon shot" height="235" width="471"></p>
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<p> </p>
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<p>Both equations are already in the "y =" format, so set them equal to each other:</p>
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<div class="so">0.15x = 2 + 0.12x - 0.002x<sup>2</sup></div>
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<p>Simplify into "= 0" format:</p>
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<div class="so">Bring all terms to left: 0.002x<sup>2</sup> + 0.15x - 0.12x - 2 = 0 </div>
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<div class="so">Simplify: 0.002x<sup>2</sup> + 0.03x - 2 = 0 </div>
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<div class="so">Multiply by 500: x<sup>2</sup> + 15x - 1000 = 0</div>
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<p>Solve the Quadratic Equation:</p>
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<div class="so">Split 15x into -25x+40x: x<sup>2</sup> -25x + 40x - 1000 = 0</div>
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<div class="so">Then: x(x-25) + 40(x-25) = 0</div>
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<div class="so">Then: (x+40)(x-25) = 0</div>
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<div class="so">x = -40 or 25</div>
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<p>The negative answer can be ignored, so <b>x = 25</b></p>
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<p>Use the linear equation to calculate matching "y" value:</p>
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<div class="so"> y = 0.15 x 25 = 3.75</div>
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<p> </p>
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<p>So the cannonball impacts the slope at <b>(25, 3.75)</b></p>
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<p> </p>
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<p>You can also find the answer graphically by using the <a href="../data/function-grapher228f.html?func1=2+0.12x-0.002x%5E2&func2=0.15x&xmin=-48&xmax=48&ymin=-32&ymax=32">Function Grapher</a>:</p>
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<p class="center"><img src="images/linear-quadratic-6.gif" alt="linear quadratic graph" height="101" width="230"></p>
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</div>
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<h2>Both Variables Squared</h2>
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<p>Sometimes BOTH terms of the quadratic can be squared:</p>
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<div class="example">
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<h3>Example: Find the points of intersection of</h3>
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<p>The circle <b>x<sup>2</sup> + y<sup>2</sup> = 25</b></p>
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<p>And the straight line <b>3y - 2x = 6</b></p>
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<p class="center"><img src="images/linear-quadratic-4.gif" alt="line 3y-2x=6 vs circle x^2+y^2=25" height="190" width="203"></p>
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<p>First put the line in "y=" format:</p>
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<div class="so">Move 2x to right hand side: 3y = 2x + 6</div>
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<div class="so">Divide by 3: y = 2x/3 + 2</div>
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<p>NOW, Instead of making the circle into "y=" format, we can use <b><a href="substitution.html">substitution</a></b> (replace "y" in the quadratic with the linear expression):</p>
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<div class="so">Put y = 2x/3 + 2 into circle equation: x<sup>2</sup> + (2x/3 + 2)<sup>2</sup> = 25 </div>
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<div class="so">Expand: x<sup>2</sup> + 4x<sup>2</sup>/9 + 2(2x/3)(2) + 2<sup>2</sup> = 25 </div>
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<div class="so">Multiply all by 9: 9x<sup>2</sup> + 4x<sup>2</sup> + 2(2x)(2)(3) + (9)(2<sup>2</sup>) = (9)(25) </div>
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<div class="so">Simplify: 13x<sup>2</sup>+ 24x + 36 = 225 </div>
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<div class="so">Subtract 225 from both sides: 13x<sup>2</sup>+ 24x - 189 = 0</div>
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<p>Now it is in standard Quadratic form, let's solve it:</p>
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<div class="so">13x<sup>2</sup>+ 24x - 189 = 0</div>
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<div class="so">Split 24x into 63x-39x: 13x<sup>2</sup>+ 63x - 39x - 189 = 0</div>
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<div class="so">Then: x(13x + 63) - 3(13x + 63) = 0</div>
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<div class="so">Then: (x - 3)(13x + 63) = 0</div>
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<div class="so">So: x = 3 or -63/13</div>
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<p> </p>
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<p>Now work out y-values:</p>
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<div class="so"> Substitute x = 3 into linear equation:
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<ul>
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<li>3y - 6 = 6</li>
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<li>3y = 12</li>
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<li>y = 4</li>
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<li>So one point is <b>(3, 4)</b></li>
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</ul>
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</div>
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<div class="so"> Substitute x = -63/13 into linear equation:
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<ul>
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<li>3y + 126/13 = 6</li>
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<li>y + 42/13 = 2</li>
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<li>y = 2 - 42/13 = 26/13 - 42/13 = -16/13</li>
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<li>So the other point is <b>(-63/13, -16/13)</b></li>
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</ul>
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</div>
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<p class="center"><img src="images/linear-quadratic-7.gif" alt="line 3y-2x=6 vs circle x^2+y^2=25" height="188" width="265"></p>
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<p> <span class="center"></span></p>
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<div class="questions">8184, 8185, 8186, 8187, 8188, 8189, 8190, 8191, 8192, 8193, 8194, 8195, 8196, 8197, 8198</div>
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<div class="related"><a href="linear-equations.html">Linear Equations</a>
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<a href="quadratic-equation.html">Quadratic Equations</a>
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<a href="systems-linear-quadratic-graphical.html">Solving Systems of Linear and Quadratic Equations Graphically</a>
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