Answers
Explanation:
On the previous page, all the fractions containing radicals (or radicals containing fractions) had denominators that cancelled off or else simplified to whole numbers. What if we get an expression where the denominator insists on staying messy?
Simplify: \mathbf{\color{green}{\sqrt{\dfrac{25}{3}\,}}}
3
25
The numerator contains a perfect square, so I can simplify this:
\sqrt{\dfrac{25}{3}\,} = \dfrac{\sqrt{25\,}}{\sqrt{3\,}}
3
25
=
3
25
= \dfrac{\sqrt{5\times 5\,}}{\sqrt{3\,}} = \dfrac{5}{\sqrt{3\,}}=
3
5×5
=
3
5
This looks very similar to the previous exercise, but this is the "wrong" answer. Why? Because the denominator contains a radical. The denominator must contain no radicals, or else it's "wrong".
(Why "wrong", in quotes? Because this issue may matter to your instructor right now, but it probably won't matter to other instructors in later classes. It's like when you were in elementary school and improper fractions were "wrong" and you had to convert everything to mixed numbers instead. But now that you're in algebra, improper fractions are fine, even preferred. Similarly, once you get to calculus or beyond, they won't be so uptight about where the radicals are.)
To get the "right" answer, I must "rationalize" the denominator. That is, I must find some way to convert the fraction into a form where the denominator has only "rational" (fractional or whole number) values. But what can I do with that radical-three? I can't take the 3 out, because I don't have a pair of threes inside the radical.
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Thinking back to those elementary-school fractions, you couldn't add the fractions unless they had the same denominators. To create these "common" denominators, you would multiply, top and bottom, by whatever the denominator needed. Anything divided by itself is just 1, and multiplying by 1 doesn't change the value of whatever you're multiplying by that 1. But multiplying that "whatever" by a strategic form of 1 could make the necessary computations possible, such as when adding fifths and sevenths:
\dfrac{2}{5} + \dfrac{3}{7}
5
2
+
7
3
= \left(\dfrac{2}{5}\right)\left(\dfrac{7}{7}\right) + \left(\dfrac{3}{7}\right)\left(\dfrac{5}{5}\right)=(
5
2
)(
7
7
)+(
7
3
)(
5
5
)
= \dfrac{14}{35} + \dfrac{15}{35} = \dfrac{29}{35}=
35
14
+
35
15
=
35
29
For the two-fifths fraction, the denominator needed a factor of 7, so I multiplied by \frac{7}{7}
7
7
, which is just 1. For the three-sevenths fraction, the denominator needed a factor of 5, so I multiplied by \frac{5}{5}
5
5
, which is just 1. We can use this same technique to rationalize radical denominators.
I could take a 3 out of the denominator of my radical fraction if I had two factors of 3 inside the radical. I can create this pair of 3's by multiplying my fraction, top and bottom, by another copy of root-three. If I multiply top and bottom by root-three, then I will have multiplied the fraction by a strategic form of 1. I won't have changed the value, but simplification will now be possible:
\dfrac{5}{\sqrt{3\,}} = \left(\dfrac{5}{\sqrt{3\,}}\right)\left(\dfrac{\sqrt{3\,}}{\sqrt{3\,}}\right)
3
5
=(
3
5
)(
3
3
)
= \dfrac{5\,\sqrt{3\,}}{\sqrt{3\,}\sqrt{3\,}}=
3
3
5
3
= \mathbf{\color{purple}{ \dfrac{5\,\sqrt{3\,}}{3} }}=
3
5
3
This last form, "five, root-three, divided by three", is the "right" answer they're looking for.