Solve. 3x2 - 7x - 2 = 0
is the answer -7 _+_ sq root 25 over 6?
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jimthompson5910 (jim_thompson5910):
the discriminant should be
d = b^2 - 4ac
d = (-7)^2 - 4(3)(-2)
d = 49 + 24
d = 73
OpenStudy (anonymous):
how would it be a positive? slightly confused
jimthompson5910 (jim_thompson5910):
(-7)^2 = 49
- 4(3)(-2) = -12*(-2) = 24
jimthompson5910 (jim_thompson5910):
so
d = (-7)^2 - 4(3)(-2)
turns into
d = 49 + 24
OpenStudy (anonymous):
oh, I see now, thank you @jim_thompson5910
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jimthompson5910 (jim_thompson5910):
np, so this means that your final answers should be
\[\Large \frac{7\pm\sqrt{73}}{6}\]
OpenStudy (anonymous):
Correct, thank you!
OpenStudy (anonymous):
Solve. 3x2 - 2x - 3 = 0
OpenStudy (anonymous):
x = sq root 80 over 3?
jimthompson5910 (jim_thompson5910):
no
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OpenStudy (anonymous):
let me revise... i made an error obviously
jimthompson5910 (jim_thompson5910):
\[\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
\[\Large x = \frac{-(-2)\pm\sqrt{(-2)^2-4(3)(-3)}}{2(3)}\]
\[\Large x = \frac{2\pm\sqrt{4-(-36)}}{6}\]
\[\Large x = \frac{2\pm\sqrt{4+36}}{6}\]
\[\Large x = \frac{2\pm\sqrt{40}}{6}\]
I'll let you finish
jimthompson5910 (jim_thompson5910):
keep in mind that \[\Large \sqrt{40}\] can be simplified