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<!-- Mirrored from www.mathsisfun.com/algebra/change-of-variables.html by HTTrack Website Copier/3.x [XR&CO'2014], Sat, 29 Oct 2022 01:02:24 GMT -->
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<title>Change of Variables</title>
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<h1 align="center">Change of Variables</h1>
<p>Sometimes &quot;changing a variable&quot; can help us solve an equation.</p>
<div class="center80">
<p>The Idea: If we can't solve it <b>here</b>, then move <b>somewhere else</b> where we can solve it, and then <b>move back</b> to the original position.</p>
</div>
<p>Like this:</p>
<p align="center"><img src="images/change-of-variable-a.svg" alt="Change of Variable" /></p> <p>These are the steps: </p>
<ul>
<div class="bigul">
<li>Replace an expression (like &quot;2x-3&quot;) with a variable (like &quot;u&quot;)</li>
<li>Solve, </li>
<li>Then put the expression (like &quot;2x-3&quot;) back into the solution (where &quot;u&quot; is).</li>
</div>
</ul>
<p>&nbsp;</p>
<h2>Example</h2>
<p>Here is a simple example: solving <b>(x+1)<sup>2</sup> &minus; 4 = 0</b>.</p>
<p align="center" class="larger">Replace &quot;x+1&quot; with &quot;u&quot; ... Solve ... Replace &quot;u&quot; with &quot;x+1&quot;:</p>
<p align="center"><img src="images/change-of-variable-b.svg" alt="Change of Variable u=x+1" /></p>
<h2>More Examples</h2>
<p>OK, we could have solved that without doing that &quot;u=x+1&quot; thing, but here is question where &quot;changing variables&quot; is very useful:</p>
<div class="example">
<h3> Example: (x<sup>2</sup>+2)<sup>2</sup> &minus; 2(x<sup>2</sup>+2) &minus; 15 = 0</h3>
<p>It could be hard to solve, but let's try a change of variables:</p>
<p>&nbsp;</p>
<p>Let <span class="large">u = x<sup>2</sup>+2</span>, then our equation becomes:</p>
<p align="center" class="larger">u<sup>2</sup> &minus; 2u &minus; 15 = 0</p>
<p>Which is a <a href="quadratic-equation.html">quadratic equation</a> that <a href="factoring-quadratics.html">factors</a> nicely into:</p>
<p align="center" class="larger">(u&minus;5)(u+3)</p>
<p>And the solutions are simply:</p>
<p align="center" class="larger"><b>u = 5</b> or <b>u = &minus;3</b></p>
<p>&nbsp;</p>
<p>But wait! We still need to turn &quot;u&quot; back into &quot;x<sup>2</sup>+2&quot;:</p>
<table border="0" align="center">
<tr>
<th align="center">First Solution</th>
</tr>
<tr>
<td align="center">u = 5</td>
</tr>
<tr>
<td align="center"><span class="hilite">x<sup>2</sup>+2</span> = 5</td>
</tr>
<tr>
<td align="center">x<sup>2</sup> = 5&minus;2 = 3</td>
</tr>
<tr>
<td align="center">x = &plusmn;&radic;3</td>
</tr>
</table>
<p>&nbsp;</p>
<table border="0" align="center">
<tr>
<th align="center">Second Solution</th>
</tr>
<tr>
<td align="center">u = &minus;3</td>
</tr>
<tr>
<td align="center"><span class="hilite">x<sup>2</sup>+2</span> = &minus;3</td>
</tr>
<tr>
<td align="center"> x<sup>2</sup> = &minus;3&minus;2</td>
</tr>
<tr>
<td align="center">x<sup>2</sup> = &plusmn;&radic;(&minus;5)</td>
</tr>
</table>
<p>The second solution is <a href="../numbers/imaginary-numbers.html">imaginary</a> (it has the square root of a negative number), so let us just use the First Solution:</p>
<p align="center" class="large">Answer: x = &plusmn;&radic;3 </p>
<p>&nbsp;</p>
<p>Check: ((&radic;3)<sup>2</sup>+2)<sup>2</sup> &minus; 2((&radic;3)<sup>2</sup>+2) &minus; 15 = = 5<sup>2</sup> &minus; 2&middot;5 &minus; 15 = 25&minus;10&minus;15 = 0<br />
Check: ((&minus;&radic;3)<sup>2</sup>+2)<sup>2</sup> &minus; 2((&minus;&radic;3)<sup>2</sup>+2) &minus; 15 = = 5<sup>2</sup> &minus; 2&middot;5 &minus; 15 = 25&minus;10&minus;15 = 0</p>
</div>
<p>&nbsp;</p>
<div class="example">
<h3> Example: 3x<sup>8</sup> + 5x<sup>4</sup> &minus; 2 = 0</h3>
<p>It sort of looks Quadratic, but it is degree 8 which could be impossible to solve.</p>
<p>But if we use:</p>
<p align="center" class="larger">u = x<sup>4</sup></p>
<p>Then it becomes:</p>
<p align="center" class="larger">3u<sup>2</sup> + 5u &minus; 2 = 0</p>
<p>Which <b>is</b> Quadratic. And <a href="../quadratic-equation-solver.html">solving it</a> gives:</p>
<p align="center" class="larger"><b>u = 1/3</b> or <b>u = &minus;2</b></p>
<p>Now put the original back again:</p>
<table border="0" align="center">
<tr>
<th align="center">First Solution</th>
</tr>
<tr>
<td align="center">u = 1/3</td>
</tr>
<tr>
<td align="center"><span class="hilite">x<sup>4</sup></span> = 1/3</td>
</tr>
<tr>
<td align="center">x = (1/3)<sup>1/4</sup></td>
</tr>
</table>
<p>&nbsp;</p>
<table border="0" align="center">
<tr>
<th align="center">Second Solution</th>
</tr>
<tr>
<td align="center">u = &minus;2</td>
</tr>
<tr>
<td align="center"><span class="hilite">x<sup>4</sup></span> = &minus;2</td>
</tr>
<tr>
<td align="center">x = (&minus;2)<sup>1/4</sup></td>
</tr>
</table>
<p align="center">&nbsp;</p>
<p align="center" class="large">Answer: x = (1/3)<sup>1/4</sup> and x = (&minus;2)<sup>1/4</sup></p>
<p>Check: <b>You</b> can check this answer!</p>
</div>
<h2>Conclusion</h2>
<p>&quot;Change of Variable&quot; can help us solve difficult questions, using the steps:
</p>
<ul>
<li>Replace an expression with a variable (like "u")
</li>
<li>Solve,
</li>
<li>Put the expression back into the solution (where "u" is)</li>
</ul>
<p>&nbsp;</p>
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<script type="text/javascript">getQ(1220, 1221, 85, 577, 578, 1219, 2255, 2256, 2257, 7124);</script>&nbsp;
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