(x + 2)(3x) = (x + 2)(6)
distribute on both sides
3x^2 + 6x = 6x + 12
good, now subtract 6x from each side
3x^2 = 12
good, now divide both sides by 3
x = 2
right?
almost, don't forget x = -2 as well x = 2 and x = -2
wait, how?
when we solve x^2 = A, we must take into account the positive and negative square roots
but if its squared doesnt it turn positive?
right, that's precisely why we need to take both positive and negative roots
can you type it out so i can see?
allow me to explain
we got down to x^2 = 4, right?
right
you found x = 2, which works when we substitute back into the equation x^2 = 4 2^2 = 4
notice what happens when we let x = -2 instead x^2 = 4 (-2)^2 = 4 4 = 4 therefore, x = -2 is also a solution.
ohhhhh..... but is that how I figure it out in every situation, or is there a way so it becomes part of the routine of solving the problem?
in general: |dw:1443665411909:dw| this is a tricky concept
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