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142 lines
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<h1 align="center">Approximate Solutions</h1>
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<p>Sometimes it is difficult to solve an equation exactly. But an approximate answer may be good enough!</p>
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<h2>What is Good Enough?</h2>
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<p>Well, that depends what you are working on!</p>
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<ul>
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<li>If you are dealing with millions of dollars then you should try to get pretty close indeed. And that might need many <a href="../rounding-numbers.html">significant digits</a>.</li>
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<li>If you are calculating how much food to buy for a party, then a small error won't matter so much. You could always buy a little extra to be sure. </li>
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<li>Or something in between</li>
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</ul>
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<p>So <b>understanding what you are working on</b> helps you know how accurate you should be. </p>
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<h2>Solving Equations</h2>
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<p>To help reduce error, when solving equations:</p>
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<ul>
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<li>first solve for <b>x = <i>something</i></b></li>
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<li><b>then</b> do any calculations</li>
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</ul>
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<p>Like this:</p>
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<div class="example">
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<h3>Example: Solve x/7 − 6.3068 + 2<span class="times">π</span> = 0 (to 3 decimal places)</h3>
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<div class="example2"></div>
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<div class="tbl">
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<div class="row"><span class="left">Start with:</span><span class="right">x/7 − 6.3068 + 2<span class="times">π</span> = 0</span></div>
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<div class="row"><span class="left">Subtract <b>−6.3068+2<span class="times">π</span></b> from both sides:</span><span class="right">x/7 = +6.3068 − 2<span class="times">π</span></span></div>
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<div class="row"><span class="left">Multiply by 7:</span><span class="right">x = 7(6.3068 − 2<span class="times">π</span>)</span></div>
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<div class="row"><span class="left">NOW do the calculations:</span><span class="right">x = 0.165</span></div>
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</div>
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</div>
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<p>Why wait until the end to do the calculations? Well, every time you do a calculation you can introduce an error. If you do this several times your errors can <b>accumulate</b> to be quite large. </p>
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<h2>Checking</h2>
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<p>If your answer is approximate, then your checking will also be approximate.</p>
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<div class="example">
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<h3>Example: Check that <span class="larger">x = 0.165</span> solves x/7 − 6.3068 + 2<span class="times">π</span> = 0</h3>
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<div class="example2"></div>
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<div class="tbl">
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<div class="row"><span class="left">Substitute 0.165 for <b>x</b>:</span><span class="right">0.165/7 − 6.3068+ 2<span class="times">π</span> = 0</span></div>
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<div class="row"><span class="left">Calculate:</span><span class="right">−0.00004 = 0</span></div>
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</div>
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<p>Not quite right, but very close.</p>
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</div>
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<h2>Graphical Estimation</h2>
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<p>You can make good approximations using graphs, particularly by using a zoom function, like on our <a href="../data/function-grapher.html">Function Grapher</a>.</p>
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<p>Here is an example:</p>
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<div class="example">
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<h3>Example: estimate the solution to x<sup>3</sup> − 2x<sup>2 </sup> − 1 = 0 (to 2 decimal places).</h3>
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<p>Solution: Plot it!</p>
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<p>Here is my first attempt. I can see it crosses through y=0 at about x=2.2</p>
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<p align="center"><img src="images/intermediate-value-d.gif" alt="graph" height="139" width="171"></p>
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<p>Let us <b>zoom in</b> there to see if we can see the crossing point better:</p>
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<p align="center"><img src="images/intermediate-value-e.gif" alt="graph" height="139" width="171"></p>
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<p>It crosses between 2.20 and 2.21 ... slightly closer to 2.21. We are asked for 2 decimal places, so our answer is:</p>
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<p> </p>
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<p class="larger" align="center">x<sup>3</sup> − 2x<sup>2 </sup> − 1 = 0 at about <b>x = 2.21</b></p>
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<p> </p>
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<p>Check: (2.21)<sup>3</sup> − 2(2.21)<sup>2 </sup> + 2 = approx 0.025, close to y=0</p>
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</div>
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<p> </p>
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<div class="questions">
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<script type="text/javascript">getQ(1212, 7180, 7181, 571, 572, 1213, 2402, 2403, 8252, 8257);</script>
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<div class="related"><a href="index.html">Algebra Index</a></div>
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