Listen to what carte said, you can basically choose +/- of the radical

but, what i meant was more general
This was simple enough to reason out, what if the numbers had been large or fractions?
There are other methods to approximate the answer, but not in college algebra

the algebra to solve this can be long and tedious so i will only show the important steps.

square both sides
7x+2 * sqrt(6x+1) * sqrt(x-4) -3 = 25

add 3 and subtract 7x on both sides
2 * sqrt(6x+1) * sqrt(x-4) = 28-7x

square both sides again
4(6x+1)(x-4) = (28-7x)^2

expand both sides
24x^2-92x-16 = 49x^2-392x+784

set equation equal to 0
-25x^2+300x-800 = 0

solve for x ( i used quadratic formula but dividing by -25 and factoring works as well)
x=4 , x=8

plug back into original equation to see if they work

sqrt(6(4) +1) + sqrt((4)-4) = 5 ..... works

sqrt(6(8) +1) + sqrt((8)-4) = 5 ... works if you take the negative number of the square root of sqrt((8)-4)
therefore you can have solution or x=4 and x=8

This post was edited by cdexswzaq on Sep 28 2014 05:19pm

Thanks for your explanation, I see where I messed up now. I did divide the formula to simplify a bit before using the quadratic formula.

Reached this point x = 12±4/2 which comes out to be 4 and 8.

I would not have guessed that playing diablo 2 when I was younger would teach me and help me understand parts of school.

lol, problem solved

This post was edited by Iffy00 on Sep 29 2014 12:14pm