SOLUTION: could someone explain to me how i should siplify sqare roots as such; 2sqrt(3)+sqrt(27)-sqrt(1/3) or sqrt(1/2)*(sqrt(8)+sqrt(2)) or sqrt(18)-sqrt(3)over sqrt(6)

Question 186279: could someone explain to me how i should siplify sqare roots as such;
2sqrt(3)+sqrt(27)-sqrt(1/3)
or
sqrt(1/2)*(sqrt(8)+sqrt(2))
or
sqrt(18)-sqrt(3)over sqrt(6)
and the last thing i need help in would be; sqrt(a)+sqrt(b) over sqrt(a)-sqrt(b)
i totally suck when it comes to things like sqrt =(
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
could someone explain to me how i should siplify sqare roots as such;
2sqrt(3)+sqrt(27)-sqrt(1/3)
=2sqrt(3) + 3sqrt(3) - sqrt(1/2)
see that? sqrt(27) = sqrt(3*9) = sqrt(9)*sqrt(3) = 3sqrt(3)
-------------
Then, sqrt(1/3)= sqrt(3/9) = sqrt(3)/sqrt(9) = sqrt(3)/3 right?
so it's 2sqrt(3) +3sqrt(3) - (1/3) sqrt(3)
= (5 - 1/3)sqrt(3)
= 14sqrt(3)/3
---------------
or
sqrt(1/2)*(sqrt(8)+sqrt(2))
Multiply the sqrt(1/2) thru
=sqrt(1/2)*sqrt(8) + sqrt(1/2)*sqrt(2)
=sqrt((1/2)*8) + sqrt((1/2)*2)
=sqrt(4) + sqrt(1)
= 2+1
=3
------------
or
sqrt(18)-sqrt(3)over sqrt(6)
Divide each term of the NUM
=sqrt(18/6) - sqrt(3/6)
=sqrt(3) - sqrt(1/2)
=sqrt(3) - sqrt(2/4)
=sqrt(3) - sqrt(2)/2
-----------------
and the last thing i need help in would be;
sqrt(a)+sqrt(b) over sqrt(a)-sqrt(b)
The thing to do with these is to multiply thru by the conjugate of the DEN
The conjugate of sqrt(a)-sqrt(b) is sqrt(a)+sqrt(b). The same 2 terms with a different sign (+ instead of -).
=(sqrt(a)+sqrt(b))*(sqrt(a)+sqrt(b))/(sqrt(a)-sqrt(b))*(sqrt(a)-sqrt(b))
It can get messy.
=(a + 2sqrt(ab) + b)/(a-b)
------------------
email me with any questions via the thank you note.
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