How do you work square root problems when there is a number outside the radical that needs to be divided out.
Example: 2(square root of 5x-1)=3(square root of 2x+4). How can you get both of the radicals by themselves in order to solve the equation?
4 (5x-1) = 9 (2x+4) you mean?
By the way be careful to check results whenever you square both sides like that. remember -1^2 = +1^2
so you can introduce answers that do not work in the original.
The two and the three are both being multiplied to the square roots that are on both sides of the equal sign. I don't know how to isolate the radicals(to get everything being added, subtracted, multiplied, or divided away) in order to square everything to get rid of the radicals. If the problem was 4+(square root of 4+3x)=6, I know that all you do is subtract the four at the beginning and you square everything. It would be 4+3x=4 and then solve from there, which this problem probably doesn't work b/c I made it up.
look
5 sqrt (25) = sqrt (625)
5^2 (25) = 625
25 * 25 = 625
625 = 625
How are you getting the five? I am not really understanding. Since the numbers 2 and 3 are being MULTIPLIED, how can you just add them instead of dividing
this is all because
[5 sqrt(25) ][5 sqrt(25)] = 25 * 25
WHERE ARE YOU GETTING THE 5 AND THE 25. HOW ARE YOU MULTIPLYING THESE RANDOM NUMBERS TOGETHER?
I just made the problem with 5 up to show what is happening
the point is that in your problem
[ 2 sqrt(5x-1) ] [ 2 sqrt (5x-1) ]
= 2*2 * (5x-1)
= 4 * (5x-1)
Perhaps say
(a * b)^2 = a^2 * b^2
whatever a and b are
my teacher told me that the only way you can square EVERYTHING was if you ISOLATE the radicals. What you're doing is something completely different.
You could have told me you made up a completely different problem.