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	<title>Comments on: FL. Algebra on the Radio</title>
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		<title>By: Stephen Morris</title>
		<link>http://mathfactor.uark.edu/2009/03/fl-algebra-on-the-radio/comment-page-1/#comment-476</link>
		<dc:creator>Stephen Morris</dc:creator>
		<pubDate>Tue, 31 Mar 2009 01:03:52 +0000</pubDate>
		<guid isPermaLink="false">http://mathfactor.uark.edu/?p=504#comment-476</guid>
		<description>This made me ask another question.
&lt;div&gt;Given a value for the product/sum can you always come up with a pair of numbers that gives you that product/sum?&lt;/div&gt;
&lt;div&gt;Is this pair unique?&lt;/div&gt;
&lt;div&gt;To put it another way: given a number, k, can you always come up with a pair of numbers, a and b, such that a+b = ab = k?&lt;/div&gt;
&lt;div&gt;Consider the expression (x-a)(x-b) = x^2 - (a+b)x + ab = x^2 - kx + k&lt;/div&gt;
&lt;div&gt;This is zero only when x=a or x=b. &lt;/div&gt;
&lt;div&gt;So a and b are the solutions to the equation x^2 - kx + k = 0.&lt;/div&gt;
&lt;div&gt;We can find a and b by the standard forumala for a quadratic which gives&lt;/div&gt;
&lt;div&gt;  a, b = {k +- sqrt(k(k-4)) }/2&lt;/div&gt;
&lt;div&gt;Now consider that k(k-4) term.  &lt;/div&gt;
&lt;div&gt;If it is negative we don&#039;t have a solution, not in real numbers anyway.&lt;/div&gt;
&lt;div&gt;If it is zero we have a solution with a = b.  This happens if k= 0 or k=4 which gives a=b=0 or a=b=2 respectively.&lt;/div&gt;
&lt;div&gt;If it is positive we have a unique solution in the real numbers. k(k-4) is positive when k&lt;0 or k&gt; 4.&lt;/div&gt;
&lt;div&gt;So we can always find a and b when k&lt;= 0 or k&gt;= 4.&lt;/div&gt;
&lt;div&gt;Blimey! Enough with the quadratics already!&lt;/div&gt;</description>
		<content:encoded><![CDATA[<p>This made me ask another question.</p>
<div>Given a value for the product/sum can you always come up with a pair of numbers that gives you that product/sum?</div>
<div>Is this pair unique?</div>
<div>To put it another way: given a number, k, can you always come up with a pair of numbers, a and b, such that a+b = ab = k?</div>
<div>Consider the expression (x-a)(x-b) = x^2 &#8211; (a+b)x + ab = x^2 &#8211; kx + k</div>
<div>This is zero only when x=a or x=b. </div>
<div>So a and b are the solutions to the equation x^2 &#8211; kx + k = 0.</div>
<div>We can find a and b by the standard forumala for a quadratic which gives</div>
<div>  a, b = {k +- sqrt(k(k-4)) }/2</div>
<div>Now consider that k(k-4) term.  </div>
<div>If it is negative we don&#8217;t have a solution, not in real numbers anyway.</div>
<div>If it is zero we have a solution with a = b.  This happens if k= 0 or k=4 which gives a=b=0 or a=b=2 respectively.</div>
<div>If it is positive we have a unique solution in the real numbers. k(k-4) is positive when k&lt;0 or k&gt; 4.</div>
<div>So we can always find a and b when k&lt;= 0 or k&gt;= 4.</div>
<div>Blimey! Enough with the quadratics already!</div>
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	<item>
		<title>By: mathphan</title>
		<link>http://mathfactor.uark.edu/2009/03/fl-algebra-on-the-radio/comment-page-1/#comment-469</link>
		<dc:creator>mathphan</dc:creator>
		<pubDate>Fri, 27 Mar 2009 17:59:43 +0000</pubDate>
		<guid isPermaLink="false">http://mathfactor.uark.edu/?p=504#comment-469</guid>
		<description>This can be solved algebraically:
[spoiler]Let the numbers be a and b
ab = a + b

Subtract b from both sides:
ab - b = a
Factor out b:
b(a - 1) = a
Divide both sides by a - 1:
b = a / (a - 1)
Basically for whatever number you give me, I subtract 1 from it and divide it into your original number.
Examples:
a = 3
b = 3/2
a = 7
b = 7/6
a = ?
b = ? / (? - 1)
Now looking at the domain of b / (b - 1), all real numbers are possible except b = 1.

Double-check:
1x = x + 1
x = x + 1
0 = 1 (no solution)

&lt;strong&gt;Answer:
&lt;/strong&gt;You can stump me with the number 1.
[/spoiler]
</description>
		<content:encoded><![CDATA[<p>This can be solved algebraically:<br />
<a href="javascript:void(null);" onclick="s_toggleDisplay(document.getElementById('SID1121746183'), this, 'Show Spoiler &#9660;', 'Hide Spoiler &#9650;');">Show Spoiler &#9660;</a></p>
<div id='SID1121746183' style='display:none;'>
Let the numbers be a and b<br />
ab = a + b</p>
<p>Subtract b from both sides:<br />
ab - b = a<br />
Factor out b:<br />
b(a &#8211; 1) = a<br />
Divide both sides by a &#8211; 1:<br />
b = a / (a &#8211; 1)<br />
Basically for whatever number you give me, I subtract 1 from it and divide it into your original number.<br />
Examples:<br />
a = 3<br />
b = 3/2<br />
a = 7<br />
b = 7/6<br />
a = ?<br />
b = ? / (? &#8211; 1)<br />
Now looking at the domain of b / (b &#8211; 1), all real numbers are possible except b = 1.</p>
<p>Double-check:<br />
1x = x + 1<br />
x = x + 1<br />
0 = 1 (no solution)</p>
<p><strong>Answer:<br />
</strong>You can stump me with the number 1.</p>
</div>
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