simplify
Answers
Answer
Sorry in advance
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
+
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.