combining radicals: need steps for r6a -3r96a -6r24a

Er... as you've posted, there's two variables (r and a) and no radicals in the problem, and so it's a matter of combining like terms. Based on your choice of subject line, I'll go out on a limb and assume that "r" is your notation for "radical." Although, that still leaves quite a bit of confusion you'll need to clear up. Is the problem:

\(\displaystyle \sqrt{6}a-3\sqrt{96}a-6\sqrt{24}a\)

or maybe \(\displaystyle \sqrt{6a}-3\sqrt{96a}-6\sqrt{24a}\)

...or something else entirely?

When you reply back, please also include any and all work you've done on this problem, even if you know it's wrong. Thank you.
 
You may use this notation for typing square roots:

sqrt(16) means the square root of 16

sqrt(x) means the square root of x

2*sqrt(121x) means the square root of 121x is multiplied by 2
 
can someone please show me the steps to combining this?

r6a -3r96a -6r24a
I bet the OP meant r6a - 3r96a- 6r24a. It boggles my mind that someone can't tell that what they wrote is unclear.

Simply find LCM(1, 3, 6), LCM( r6, r96, r24) and LCM(a, a, a).
 
Actually I reconsider my position after rereading ksdhart2's post and now think that the way ksdhart assumption is correct.
 
rootof(6a) - 3*rootof(96a) - 6*rootof(24a)

rootof(6a) - 3*rootof(16*6*a) - 6*rootof(4*6*a)

rootof(6a) - 3*4*rootof(6a) - 6*2*rootof(6a)

rootof(6a) - 12*rootof(6a) - 12*rootof(6a)

-23*rootof(6a)
 
combine radicals

Er... as you've posted, there's two variables (r and a) and no radicals in the problem, and so it's a matter of combining like terms. Based on your choice of subject line, I'll go out on a limb and assume that "r" is your notation for "radical." Although, that still leaves quite a bit of confusion you'll need to clear up. Is the problem:

\(\displaystyle \sqrt{6}a-3\sqrt{96}a-6\sqrt{24}a\)

or maybe \(\displaystyle \sqrt{6a}-3\sqrt{96a}-6\sqrt{24a}\)

...or something else entirely?

When you reply back, please also include any and all work you've done on this problem, even if you know it's wrong. Thank you.
Could someone show me the steps to combine this?
\sqrt{6}a-3\sqrt{96}a-6\sqrt{24}a
 
Could someone show me the steps to combine this?
\(\displaystyle \sqrt{6}a-3\sqrt{96}a-6\sqrt{24}a\)

\(\displaystyle \sqrt{96}=\sqrt{16\cdot 6}=4\sqrt{6}\)
\(\displaystyle \sqrt{24}=\sqrt{4\cdot 6}=2\sqrt{6}\)
\(\displaystyle 1-12-12=-23\)
\(\displaystyle \begin{align*}\sqrt{6}a-3\sqrt{96}a-6\sqrt{24}a &=\sqrt{6}a-12\sqrt{6}a-12\sqrt{6}a \\&=-23\sqrt{6}\end{align*}\)
 
\(\displaystyle \sqrt{96}=\sqrt{16\cdot 6}=4\sqrt{6}\)
\(\displaystyle \sqrt{24}=\sqrt{4\cdot 6}=2\sqrt{6}\)
\(\displaystyle 1-12-12=-23\)
\(\displaystyle \begin{align*}\sqrt{6}a-3\sqrt{96}a-6\sqrt{24}a &=\sqrt{6}a-12\sqrt{6}a-12\sqrt{6}a \\&=-23\sqrt{6}\end{align*}\)
Shouldn't there be an "a" in that last answer?
 
Shouldn't there be an "a" in that last answer [due to typo/oversight]?

Let's just make a definitive statement. You know there is one, so just instead of asking if there is one (read: beating around the bush),
just state the correction.

\(\displaystyle \cdot\)
\(\displaystyle \cdot\)
\(\displaystyle \cdot\)

\(\displaystyle \begin{align*}\sqrt{6}a-3\sqrt{96}a-6\sqrt{24}a &=\sqrt{6}a-12\sqrt{6}a-12\sqrt{6}a \\&=-23\sqrt{6}a\end{align*}\)
 
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