Question

Can someone explain this to me because I am confused.

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Answer 1

\[\sqrt{a} \times \sqrt{b} = \sqrt{ab}\]


we have:

\[5*\sqrt{25}*\sqrt{d}*2d*\sqrt{3}\]

Answer 2

No.. I Don't think it would be. \[5\sqrt{25d} * 2d \sqrt{3}\rightarrow 5 * 5\sqrt{d} * 2d \sqrt{3}\]

Answer 3

\[25\sqrt{d} *2d \sqrt{3}\]

Answer 4

same thing i wrote :) but you got it

Answer 5

But that's not an answer :(

Answer 6

now you can multiply 25 and 2d as well as sqrt(d) and sqrt(3)

Answer 7

But the roots aren't like terms...

Answer 8

you know that
\[ 25\sqrt{d} *2d \sqrt{3} \] means the same as
\[ 25\cdot 2d \cdot \sqrt{d} \cdot \sqrt{3} \]
(change the order of multiplying)
or
\[ 50 d \sqrt{3d }\]

Answer 9

notice you use the idea
\[ \sqrt{a \cdot b}= \sqrt{a} \sqrt{b} \]
when you simplified
\[ \sqrt{25 d}= \sqrt{25} \sqrt{d} \]

Answer 10

in other words , you can switch between
\[ \sqrt{3} \sqrt{d} \text{ and } \sqrt{3d} \]

Answer 11

**But the roots aren't like terms...

that is not the "rule" we use this problem.

Answer 12

IS that only the rule when adding and subtracting?

Answer 13

yes, if you mean, for example,
\[ \sqrt{3}+ 2\sqrt{3} = 3 \sqrt{3} \]

the way to remember this is remember that sqrt(3) is a number and you can factor it out
\[ \sqrt{3}+ 2\sqrt{3} = (1+2)\sqrt{3} \]

Answer 14

thnx :)

Answer 15

I was trying to come up with a good way to explain why
\[ \sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b} \]
but the best I have is an example that it works (which is not a proof)
\[\sqrt{4}\sqrt{9}= 2\cdot 3=6 \\ \sqrt{36} = 6 \]

Answer 16

I understand what you're saying :)