Can someone show me how to solve these types of problems. i have two that are driving me nuts.as of now.
Problem #1
Simplify:
2radical(12) + 4 radical (27)
Problem #2
evaluate if possible the following
i am going to describe it
in the index it has a 3 then a radical and for the radicand it has negative 27
can't help sorry
I hope I have understood the question.
2radical(l2)=2radical(2x2x3)
2radical(2x2x3)=2radical(2x2)x2radical(3)=2x(2radical(3))
4radical(27)= 2radical(3x2radical(3))
=2radical(3)x4radical(3)
adding the two
2x(2radical(3))+2radical(3)x4radical(3)=
taking 2radical(3) out as a factor
=2radical(3)x(2+4radical(3))
Does this help?
for 3radical(-27)
3radical(27) would be 3 [3x3x3=27)
3radical(-27) would be -3 (-3x-3x-3=-27)
To simplify the expression 2√12 + 4√27, we can start by finding the prime factorization of each radicand.
For √12:
12 = 2 x 2 x 3
Thus, √12 = 2√3
For √27:
27 = 3 x 3 x 3
Thus, √27 = 3√3
Now we can substitute these values back into the original expression:
2√12 + 4√27 = 2(2√3) + 4(3√3)
Next, we can simplify each term:
2(2√3) = 4√3
4(3√3) = 12√3
Combining the two simplified terms:
4√3 + 12√3 = (4 + 12)√3 = 16√3
Therefore, the simplified expression becomes 16√3.
For the second problem, you're trying to evaluate √(-27) with an index of 3.
To evaluate this, first note that the cube root of -27 is -3 because -3 x -3 x -3 = -27.
Therefore, √(-27) = -3.