Hello, cvandoren!
\(\displaystyle 8\sqrt{5}\,-\,2\sqrt{45}\)
Heres what I got so far:
\(\displaystyle 8\sqrt{5}\) can be reduced to \(\displaystyle 2\sqrt{5}\) . . . really??
How did you change the "8" to a "2" ?
Evidently, you don't understand Simplifying Radicals.
The basic rule we use is: \(\displaystyle \;\sqrt{a\cdot b}\:=\:\sqrt{a}\cdot\sqrt{b}\)
. . The square root of a product may be "split" into two square roots.
Example: Simplify \(\displaystyle \sqrt{45}\)
Try to factor the 45 into two parts, where one part is a square number.
. . We find that \(\displaystyle 45\:=\:5\,\times\,9\) . . . and 9 is a square.
So we have: \(\displaystyle \;\sqrt{45}\;=\;\sqrt{9\cdot5}\) . . . I always put the square "in front"
. . Then we can "split" it: \(\displaystyle \sqrt{9\cdot5}\;=\;\sqrt{9}\cdot\sqrt{5}\)
We know that \(\displaystyle \sqrt{9}\,=\,3\) ... and we're stuck with the \(\displaystyle \sqrt{5}\).
. . So we have: \(\displaystyle \;3\sqrt{5}\)
Back to the original problem: \(\displaystyle \;8\sqrt{5}\,-\,2\sqrt{45}\)
We just learned that \(\displaystyle \sqrt{45}\,=\,3\sqrt{5}\), right?
. . So we have: \(\displaystyle \;8\sqrt{5}\,-\,2\cdot3\sqrt{5} \;= \;8\sqrt{5}\,-\,6\sqrt{5}\)
Can we combine them?
.Yes, it's exactly like \(\displaystyle 8x\,-\,6x\) ("like" terms)
Answer: \(\displaystyle \;2\sqrt{5}\)
