Mathematics
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OpenStudy (joshoyen):
What is the zero of 2x^2 + 7?
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OpenStudy (joshoyen):
Is it -2^2/7?
OpenStudy (agent0smith):
Set it equal to zero, then solve for x.
OpenStudy (shadowlegendx):
\[2x^2 + 7 = 0\]
Solve for X as smith said, simple algebra.
Start by subtracting 7
OpenStudy (joshoyen):
2x^2 = -7
OpenStudy (joshoyen):
ohh it's -7/2x^2
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OpenStudy (joshoyen):
right?
OpenStudy (joshoyen):
@ShadowLegendX
OpenStudy (joshoyen):
@agent0smith
OpenStudy (shadowlegendx):
\[2x^2 + 7 = 0\]
Solve for X as smith said, simple algebra.
Start by subtracting 7
OpenStudy (agent0smith):
2x^2 = -7
Up to here you're right, but you did something very wrong from there. Just divide both sides by 2, NOT 2x^2.
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OpenStudy (joshoyen):
-7/2
OpenStudy (agent0smith):
Yes. So what is your equation now? It's not just -7/2, that's not an equation.
OpenStudy (joshoyen):
x-7/2 ?
OpenStudy (joshoyen):
woops
OpenStudy (agent0smith):
Your equation was
2x^2 = -7
Then you divided both sides by 2.
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OpenStudy (joshoyen):
Ohh. x^2 = -7/2
OpenStudy (agent0smith):
Yes. Now you can solve by taking the square root of both sides.
OpenStudy (joshoyen):
Hmm, i'm not sure on how to do that
OpenStudy (mathstudent55):
If \(x^2 = k\), then \(x = \pm \sqrt k\).