Question 178082
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1 divided by 7 times the square root of 5 minus 3 times the square root of 3

Depends.

What you wrote can be interpreted:

*[tex \Large \frac {1} {7 sqrt{5} - 3 sqrt{3}}]

or

*[tex \Large \frac {1} {7 sqrt{5}} - 3 sqrt{3}]

Second way first, because that one is easier.  You need to find a multiplier that will change *[tex \Large 7 sqrt {5}] into a rational number.   *[tex \Large sqrt{5}] will do this quite tidily.

The process is to multiply the fractional part of your expression by 1 in the form of *[tex \Large \frac {sqrt{5}}{sqrt{5}}], thus:

*[tex \Large (\frac {1} {7 sqrt{5}})( \frac {sqrt{5}}{sqrt{5}} ) - 3 sqrt{3}]

yielding:

*[tex \Large \frac {sqrt{5}}{7 * 5} - 3 sqrt{3}]

then apply the LCD of 35 to get:

*[tex \Large \frac {sqrt{5} - 135 sqrt{3}}{35}]

The other interpretation is a bit trickier, but not too.

*[tex \Large \frac {1} {7 sqrt{5} - 3 sqrt{3}}]

You can't simply square the denominator because you would still end up with a two term expression one term of which would still be irrational.  The trick is to remember the factorization of the difference of two squares:

*[tex \Large a^2 - b^2 = (a + b)(a - b)]

*[tex \Large (a + b) \text {  and  } \math (a - b)] are called a conjugate pair. The thing is, if you multiply a binomial by its conjugate, you get the difference of two squares.

The denominator of your expression is a binomial *[tex \Large 7 sqrt{5} - 3 sqrt {3}] whose conjugate is *[tex \Large 7 sqrt{5} + 3 sqrt {3}]

The process is to multiply the entire original fraction by 1 in the form of the conjugate of the denominator over itself, thus:

*[tex \Large \left(\frac {1} {7 sqrt{5} - 3 sqrt{3}} \right)\left(\frac {7 sqrt{5} + 3 sqrt{3}}{7 sqrt{5} + 3 sqrt{3}}\right)=\left(\frac{7 sqrt{5} + 3 sqrt{3}}{245 - 27}\right)=\left(\frac{7 sqrt{5} + 3 sqrt{3}}{218}\right)]

Yes, the answer is uglier than a mud fence, but because it has a rational number for a denominator, it is simpler than the original by definition.

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