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Mathematics 20 Online
OpenStudy (joshoyen):

What is the zero of 2x^2 + 7?

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

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.

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.

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 \(x = \pm \sqrt k\).

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