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What is the simplified sum of √3+√12=

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Hint √12 = √ (4 * 3 ) = ....?
the sqrt(12) = 2 sqrt(3) so sqrt(3) + 2 sqrt(3) is your solution
It can be simpler.

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Other answers:

sqrt(3) is IMPLIED 1*sqrt(3) . What is 1 + 2?
All i care about is her understanding the methods on how to get it, regardless of how to do it
o_O Very good.
now I am confused
just follow the method I provided
You didn't provide any method, you provided a solution ..... :/ Here, you can rewrite sqrt(12) like @Chlorophyll said.
sqrt(12) = the terms sqrt(4) * sqrt(3) sqrt(4) = 2 which stays on the outside of the radical 2sqrt(3) + sqrt(3) which is the solution
either works 3sqrt(3) or sqrt(3) = 2sqrt(3)
Right, but one is simplified all of the way, and one is not. If you are providing answers to a computer (all too common), it will complain that it is not completely simplified. It's important to see how they add like that.
no answers to computers will accept both such as webassign or webwork
as long as the programs are accurate
I've tutored for years, and I can assure that this is not the case whatsoever. You can continue being indignant though.
rannsan r u clear?
yes thank you a lot I have another one for yall....................sorry for the yall I am in the south.
square root 3 value 1.732 so 1.732+2(1.732)=5.196
@rannsan We explain so that you can understand and work by YOURSELF :)
Do not do this.
That answer will be an approximation (with round off error). The exact answer is that as has been shown.
@SOMU.kvp You should learn from the others' solution!
what mr.chlorophy?
Decimal number isn't allowed to solve these types!
They are approximations. You can't always just plug and chug to get numbers out. For example. Pi = 3.14...this is an approximation and is far from exact. Pi, is exact. Similarly sqrt(3)... 1.73 is an approximation. sqrt(3) is exact. When you have problems that grow to a page or fill a chalk board, doing rounding like this will give you VERY wrong answers especially if you do it early in the problem. It's called error propagation. Google it.

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