Mathematics
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OpenStudy (anonymous):
Solve equation x(x-2)(x+3)=0
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jimthompson5910 (jim_thompson5910):
Hint:
if AB = 0, then A = 0 or B = 0
jimthompson5910 (jim_thompson5910):
So this means that if
x(x-2)(x+3)=0
then
x=0, x-2=0, or x+3=0
jimthompson5910 (jim_thompson5910):
Solve each equation for x to find your three answers
OpenStudy (anonymous):
I got it to (x^2-2x)(x+3)=0
jimthompson5910 (jim_thompson5910):
no need to do that
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OpenStudy (anonymous):
I just don't know how to distribute it
jimthompson5910 (jim_thompson5910):
it's valid of course, but you're going backwards
OpenStudy (anonymous):
that's the way my school wants me to do it though
jimthompson5910 (jim_thompson5910):
really? that's odd...
OpenStudy (anonymous):
oh wait
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jimthompson5910 (jim_thompson5910):
do you see what I wrote above? that's the best way to do it
OpenStudy (anonymous):
yea, I understand
OpenStudy (anonymous):
how do you get the result then?
jimthompson5910 (jim_thompson5910):
alright great
jimthompson5910 (jim_thompson5910):
solve each equation for x
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OpenStudy (anonymous):
I need a solution set
OpenStudy (anonymous):
ok
jimthompson5910 (jim_thompson5910):
x-2=0
x = ???
jimthompson5910 (jim_thompson5910):
x+3=0
x = ???
OpenStudy (anonymous):
2
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jimthompson5910 (jim_thompson5910):
good, and what else
OpenStudy (anonymous):
aren't I only suppose to get 2 answers for a solution set?
OpenStudy (anonymous):
wait wait wait
jimthompson5910 (jim_thompson5910):
no, in this case, you'll get 3
OpenStudy (anonymous):
if I get it to (x^2-2x)(x+3)=0, can I just divide the first one by 2 and get (x-2)(x+3)=0?
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jimthompson5910 (jim_thompson5910):
no need to do that though
OpenStudy (anonymous):
so the answer is x=2 and x=-3?
jimthompson5910 (jim_thompson5910):
there's one more
OpenStudy (anonymous):
0?
jimthompson5910 (jim_thompson5910):
bingo
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jimthompson5910 (jim_thompson5910):
x(x-2)(x+3)=0
x=0, x-2=0, or x+3=0
x=0, x=2, or x = -3
OpenStudy (anonymous):
woo!!
OpenStudy (anonymous):
thank you :3
jimthompson5910 (jim_thompson5910):
yw