Week 4 – Precalc 11

One of the things we learned this week in Precalc 11 is Dividing Radicals. Basically dividing a radical by a radical, the main goal is to turn the denomination into a rational number.
Ex.
\frac{\sqrt{10}}{\sqrt{5}}
This one is actually pretty easy, you can just divide 10 by 5 so the answer would be \sqrt{2}

In the case of \frac{\sqrt{20}}{\sqrt{3}} you can’t divide 20 by 3
we can multiply it by \frac{\sqrt{3}}{\sqrt{3}} so the denominator is a perfect square.
=\frac{\sqrt{20}}{\sqrt{3}}x\frac{\sqrt{3}}{\sqrt{3}}
=\frac{\sqrt{60}}{\sqrt{9}}
then we can simply it to
=\frac{2\sqrt{15}}{3}

For binomial radicals in the denominator, we have to use conjugates (a+b)(a-b) to square a and b term.
Ex.
\frac{\sqrt{4}}{\sqrt{3}-\sqrt{5}}
we have to use the conjugates of \sqrt{3}-\sqrt{5} which is \sqrt{3}+\sqrt{5}
=\frac{\sqrt{4}}{\sqrt{3}-\sqrt{5}}x\frac{\sqrt{3}+\sqrt{5}}{\sqrt{3}+\sqrt{5}}
=\frac{\sqrt{12}+\sqrt{20}}{\sqrt{9}-\sqrt{25}}
=\frac{2\sqrt{3}+2\sqrt{5}}{3-5}
=\frac{2\sqrt{3}+2\sqrt{5}}{-2}

More examples

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