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<h1 align="center">Solving Radical Equations</h1>
<p align="center"><i>How to solve equations with square roots, cube roots, etc.</i></p>
<h2>Radical Equations</h2>
<table border="0" align="center">
<tr>
<td><img src="../images/square-root-symbol.gif" alt="radical" width="31" height="56" /></td>
<td>&nbsp;</td>
<td>A Radical Equation is an equation with a <a href="../square-root.html">square root</a> or <a href="../numbers/cube-root.html">cube root</a>, etc.</td>
</tr>
</table>
<h2>Solving Radical Equations</h2>
<p>We can get rid of a square root by squaring. (Or cube roots by cubing, etc)</p>
<div class="center80">
<p>But Warning: this can sometimes create &quot;solutions&quot; which don't actually work when we put them into the original equation. So we need to Check!</p>
</div>
<p>&nbsp;</p>
<p>Follow these steps:</p>
<ul>
<li>isolate the square root on one side of the equation</li>
<li>square both sides of the equation</li>
</ul>
<p>Then continue with our solution!</p>
<div class="example">
<h3>Example: solve &radic;(2x+9) &minus; 5 = 0</h3>
<div class="tbl">
<div class="row"><span class="left">isolate the square root:</span><span class="right">&radic;(2x+9) = 5</span></div>
<div class="row"><span class="left">square both sides:</span><span class="right">2x+9 = 25</span></div>
</div>
<p>Now it should be easier to solve!</p>
<div class="tbl">
<div class="row"><span class="left">Move 9 to right:</span><span class="right"> 2x = 25 &minus; 9 = 16</span></div>
<div class="row"><span class="left">Divide by 2:</span><span class="right">x = 16/2 = 8</span></div>
<div class="row"><span class="left">Answer:</span><span class="right">x = 8</span></div>
</div>
<p>Check: &radic;(2&middot;8+9) &minus; 5 = &radic;(25) &minus; 5 = 5 &minus; 5 = 0</p>
</div>
<p>That one worked perfectly.</p>
<h2>More Than One Square Root</h2>
<p>What if there are two or more square roots? Easy! Just repeat the process for each one.</p>
<p>It will <b>take longer</b> (lots more steps) ... but nothing too hard.</p>
<div class="example">
<h3>Example: solve &radic;(2x&minus;5) &minus; &radic;(x&minus;1) = 1</h3>
<div class="tbl">
<div class="row"><span class="left">isolate one of the square roots:</span><span class="right">&radic;(2x&minus;5) = 1 + &radic;(x&minus;1)</span></div>
<div class="row"><span class="left">square both sides:</span><span class="right">2x&minus;5 = (1 + &radic;(x&minus;1))<sup>2</sup></span></div>
</div>
<p align="center" class="larger">We have removed one square root.</p>
<p align="center" class="larger">&nbsp;</p>
<div class="tbl">
<div class="row"><span class="left"> expand right hand side:</span><span class="right">2x&minus;5 = 1 + 2&radic;(x&minus;1) + (x&minus;1)</span></div>
<div class="row"><span class="left">simplify:</span><span class="right">2x&minus;5 = 2&radic;(x&minus;1) + x</span></div>
<div class="row"><span class="left">subtract x from both sides:</span><span class="right">x&minus;5 = 2&radic;(x&minus;1)</span></div>
</div>
<p>Now do the &quot;square root&quot; thing again:</p>
<div class="tbl">
<div class="row"><span class="left">isolate the square root:</span><span class="right">&radic;(x&minus;1) = (x&minus;5)/2</span></div>
<div class="row"><span class="left">square both sides:</span><span class="right">x&minus;1 = ((x&minus;5)/2)<sup>2</sup></span></div>
</div>
<p align="center" class="larger">We have now successfully removed both square roots.</p>
<p>&nbsp;</p>
<p>Let us continue on with the solution.</p>
<div class="tbl">
<div class="row"><span class="left">Expand right hand side:</span><span class="right">x&minus;1 = (x<sup>2</sup> &minus; 10x + 25)/4</span></div>
</div>
<p class="center larger">It is a Quadratic Equation! So let us put it in standard form.</p>
<div class="tbl">
<div class="row"><span class="left">Multiply by 4 to remove division:</span><span class="right">4x&minus;4 = x<sup>2</sup> &minus; 10x + 25</span></div>
<div class="row"><span class="left">Bring all to left:</span><span class="right">4x &minus; 4 &minus; x<sup>2</sup> + 10x &minus; 25 = 0</span></div>
<div class="row"><span class="left">Combine like terms:</span><span class="right">&minus;x<sup>2</sup> + 14x &minus; 29 = 0</span></div>
<div class="row"><span class="left">Swap all signs:</span><span class="right">x<sup>2</sup> &minus; 14x + 29 = 0</span></div>
</div>
<p>&nbsp;</p>
<p>Using the <a href="../quadratic-equation-solver.html">Quadratic Formula</a> (a=1, b=&minus;14, c=29) gives the solutions: </p>
<p align="center"><span class="large">2.53 </span>and<span class="large"> 11.47</span> (to 2 decimal places)</p>
<p>Let us check the solutions:</p>
<p><span class="large">2.53</span>: &radic;(2&times;2.53&minus;5) &minus; &radic;(2.53&minus;1) &asymp; <b>&minus;1</b> Oops! Should be plus 1. <img src="../images/style/no.svg" alt="not" /></p>
<p><span class="large">11.47</span>: &radic;(2&times;11.47&minus;5) &minus; &radic;(11.47&minus;1) &asymp; <b>1</b> Yes that one works. <img src="../images/style/yes.svg" alt="yes" /></p>
<p>There is <b>really only one solution</b>: </p>
<p>&nbsp;</p>
<p align="center"><span class="large">Answer: 11.47</span> (to 2 decimal places)</p>
</div>
<p>See? This method <b>can</b> sometimes produce solutions that don't really work!</p>
<p>The root that seemed to work, but wasn't right when we checked it, is called an <i><b>&quot;Extraneous Root&quot;</b></i><b></b></p>
<div class="center80">
<p align="center">So: Checking is important.</p>
</div>
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