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Mathematics 15 Online
OpenStudy (theloshua):

(x + 2)(3x) = (x + 2)(6)

Vocaloid (vocaloid):

distribute on both sides

OpenStudy (theloshua):

3x^2 + 6x = 6x + 12

Vocaloid (vocaloid):

good, now subtract 6x from each side

OpenStudy (theloshua):

3x^2 = 12

Vocaloid (vocaloid):

good, now divide both sides by 3

OpenStudy (theloshua):

x = 2

OpenStudy (theloshua):

right?

Vocaloid (vocaloid):

almost, don't forget x = -2 as well x = 2 and x = -2

OpenStudy (theloshua):

wait, how?

Vocaloid (vocaloid):

when we solve x^2 = A, we must take into account the positive and negative square roots

OpenStudy (theloshua):

but if its squared doesnt it turn positive?

Vocaloid (vocaloid):

right, that's precisely why we need to take both positive and negative roots

OpenStudy (theloshua):

can you type it out so i can see?

Vocaloid (vocaloid):

allow me to explain

Vocaloid (vocaloid):

we got down to x^2 = 4, right?

OpenStudy (theloshua):

right

Vocaloid (vocaloid):

you found x = 2, which works when we substitute back into the equation x^2 = 4 2^2 = 4

Vocaloid (vocaloid):

notice what happens when we let x = -2 instead x^2 = 4 (-2)^2 = 4 4 = 4 therefore, x = -2 is also a solution.

OpenStudy (theloshua):

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?

Vocaloid (vocaloid):

in general: |dw:1443665411909:dw| this is a tricky concept

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