<PRE><B>Question 277548</B>
{{{(sqrt(a) + sqrt(b))/( sqrt(a) -sqrt(b))}}}
When working with fractions you know that your answer should be reduced and if the fraction is improper it should be converted to a mixed number. When working with square roots:&lt;ul&gt;&lt;li&gt;The square roots should be in &quot;reduced&quot; form. This means that there should be no perfect square factors in the radicand (the expression within the radical).&lt;/li&gt;&lt;li&gt;Denominators must be rational. This means:&lt;ul&gt;&lt;li&gt;No fractions within a radical&lt;/li&gt;&lt;li&gt;No square roots in any denominators.&lt;/li&gt;&lt;/ul&gt;&lt;/li&gt;&lt;/ul&gt;
Your square roots have no perfect square factors (yet). But you do have square roots in a denominator so we need to rationalize it.&lt;br&gt;
Your denominator has two terms and rationalizing two-term denominators is trickier that you might think. You might think to square the numerator and denominator. But there are two things wrong with that:&lt;ol&gt;&lt;li&gt;Squaring the numerator and denominator is not a valid operation. The new fraction you get will not be the same as the original equation.&lt;/li&gt;&lt;li&gt;Even if squaring the numerator and denominator was allowed, it would still not rationalize the denominator. This is true because {{{(p - q)^2}}} is NOT equal to {{{p^2 - q^2}}}! Exponents do NOT distribute. {{{(p - q)^2}}} means {{{(p - q)*(p-q)}}} and if you multiply it out properly (using FOIL or otherwise), you get {{{p^2 -2pq + q^2}}}.&lt;/li&gt;&lt;/ol&gt;
So how do we rationalize two-term denominators? Well I hope the expression {{{p^2 - q^2}}} provides a hint. It should look familiar. There is a pattern used in factoring that involves an expression like this:
{{{(p + q)(p - q) = p^2 - q^2}}}
This points us toward an answer. It tells us that we can take a two-term expression and turn it into a two-term expression &lt;i&gt;of perfect squares&lt;/i&gt;. If we have a two-term expression with a &quot;+&quot; between them, like (p+q), we can get perfect squares if we multiply it by (p-q). And if we have a two-term expression with a &quot;-&quot; between them, like (p-q) and like your denominator, we can get perfect squares by multiplying by (p+q).&lt;br&gt;
So to rationalize your denominator we need to multiply it by {{{sqrt(a)+sqrt(b)}}}. And if we multiply the denominator we have also have to multiply the numerator by the same thing:
{{{((sqrt(a) + sqrt(b))/( sqrt(a) -sqrt(b)))((sqrt(a)+sqrt(b))/(sqrt(a)+sqrt(b)))}}}
The multiplication in the denominator is easy. The pattern tells us what we get. In the numerator we can use FOIL, the pattern for {{{(p+q)^2}}} or other proper techniques:
{{{((sqrt(a))^2 + 2sqrt(a)sqrt(b) + (sqrt(b))^2)/((sqrt(a))^2 - (sqrt(b))^2)}}}
which simplifies to:
{{{(a + 2sqrt(ab) + b)/(a - b)}}}
There are no square roots in the denominator and the only square root remaining has no perfect square factors in its radicand. And the fraction will not reduce so we are finished.</PRE>