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Mathematics 20 Online
OpenStudy (theloshua):

(x + 2)(x - 3) = (x - 3)^2

Nnesha (nnesha):

solve for x ??

OpenStudy (theloshua):

yes

OpenStudy (theloshua):

factor?

OpenStudy (solomonzelman):

divide by (x-3) 9and that gives you x-3=0 solution right of, leaving you with x+2=x-3 which is ....

Nnesha (nnesha):

alright (x-3)^2 is same as (x-3)(x-3 )\[\huge\rm (x+2)(x-3)=(x-3)(x-3)\] you can divide both sides by x-3 and then solve for x

OpenStudy (solomonzelman):

yes, but again as I noted, when you divide by x-3, you will have that x-3=0 is also a solution.

OpenStudy (solomonzelman):

(to check 0=0)

hero (hero):

You don't have to divide by x-3 By observation, x+2=x-3

hero (hero):

And solving for x doesn't seem to yield a solution.

OpenStudy (solomonzelman):

x=3

OpenStudy (theloshua):

Im confused ppl.. walk me through it?

OpenStudy (solomonzelman):

that is from dividing by x-3, and I am sure you would get that solution if you were to just solve it in another way.

hero (hero):

Xes x=3 is a solution but you don't need to divide anything to get that

Nnesha (nnesha):

ahh i see. you can distribute the parentheses FOL it (x-3)(x-3) and (x-3)(x+2)

Nnesha (nnesha):

foil* familiar with the foil method ?

Nnesha (nnesha):

distribute and then solve for x u will get the correct answer.

hero (hero):

Nope, you don't need foil EITHER

hero (hero):

hero (hero):

That is the correct solution.

Nnesha (nnesha):

why_not. yes,you_can_if_you_want.

Nnesha (nnesha):

you_will_get_the_same_answer. but_yeah_the_other_one_is_Easy_than_the_foil.

hero (hero):

Just teasin

Nnesha (nnesha):

cool..

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