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OpenStudy (kimberlyevens):
OpenStudy (kimberlyevens):
@jhonyy9
jhonyy9 (jhonyy9):
2x^2 +2x -2 = 0 so what you see there common in every 3 terms ?
OpenStudy (kimberlyevens):
Um there is a 2
jhonyy9 (jhonyy9):
yes so factoriz out the 2 and so what will get ?
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OpenStudy (kimberlyevens):
factorize out the 2?
jhonyy9 (jhonyy9):
yes from every 3 terms and so what will get ?
jhonyy9 (jhonyy9):
2(---)=0
jhonyy9 (jhonyy9):
what will be inside parantheses ?
jhonyy9 (jhonyy9):
one quadratic - do you can write me this quadratic please ?
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jhonyy9 (jhonyy9):
2(x^2 +x -1) =0
OpenStudy (kimberlyevens):
(2x^2 + 2x)
OpenStudy (kimberlyevens):
well i was way off
jhonyy9 (jhonyy9):
not what you wrote than you factoriz out from every 3 terms the 2
jhonyy9 (jhonyy9):
so now you need to solve this quadratic - do you know it how ?
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OpenStudy (kimberlyevens):
no
jhonyy9 (jhonyy9):
using discriminant
jhonyy9 (jhonyy9):
x^2 +x -1=0
what is formula of discriminant ?
jhonyy9 (jhonyy9):
D = b^2 -4ac
so D in this case how many will be ?
jhonyy9 (jhonyy9):
are you here ?
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jhonyy9 (jhonyy9):
D = 1 -4(1)(-1) = ?
jhonyy9 (jhonyy9):
b^2 =1
jhonyy9 (jhonyy9):
D = 1+4= ?
OpenStudy (anonymous):
answer is the second one
OpenStudy (t__prodigy):
The answer is b
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OpenStudy (t__prodigy):
divide thru by 2. So the equation is x^2 + x -1 = 0. Invariably, a=1, b=1, c=-1
Using the general formular -b + sqrt(b^2-4ac)/2a or -b-sqrt(b^2-4ac)/2a
OpenStudy (t__prodigy):
substitute the values of a b and c...then you are good to go