Question 182446
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Rationalize the denominator:


*[tex \LARGE \text{          }\math \frac {7}{1 - sqrt{3}}]


Rationalizing a denominator means to eliminate any irrational numbers in the denominator, such as the *[tex \Large sqrt{3}] in your example.


It won't do any good to square the denominator because you would still end up with an irrational term, so you need to remember the 'difference of two squares' factorization:


*[tex \LARGE \text{          }\math (a + b)(a - b) = a^2 - b^2]


So you can see that if you multiply a binomial by its conjugate (formed by using the same terms but changing the sign between them) you get the difference of the two squared terms.


In your case, the conjugate of your denominator, *[tex \Large 1 - sqrt {3}] is *[tex \Large 1 + sqrt {3}].


Now we can't simply multiply the denominator by its conjugate, because that would change the value of the fraction.  What we need to do is multiply the <i><b>entire fraction</b></i> by 1 in the form of:


*[tex \LARGE \text{          }\math \frac {1 + sqrt {3}}{1 + sqrt {3}} ],


like this:


*[tex \LARGE \text{          }\math \left( \frac {7}{1 - sqrt{3}} \right)\left(\frac { 1 + sqrt {3} }{1 + sqrt {3}}\right) = \frac {7 (1 + sqrt {3}) }{(1 - sqrt{3})(1 + sqrt {3})} = \frac {7 + 7sqrt {3} }{1 - 3} = \frac {7 + 7sqrt {3} }{-2}]


Now you have a nice rational number for a denominator, and therefore the expression is simpler than what you started with; the fact that the numerator is uglier than a mud fence notwithstanding.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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