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Mathematics 17 Online
OpenStudy (anonymous):

Which of the following is a solution of x2 + 5x = -2 5 plus or minus the square root of 33 divided by two. 5 plus or minus the square root of 17 divided by two negative 5 plus or minus the square root of 33 divided by two. negative 5 plus or minus the square root of 17 divided by two.

OpenStudy (nfcfox):

You can take the x out of the equation.\[x(x+5)=-2 \] \[x=-2 and x+5=-2 x=-7\]

OpenStudy (nfcfox):

I don't know where you are getting square roots from.

OpenStudy (nfcfox):

Im assuming the x2 is x^2

OpenStudy (anonymous):

\[\frac{ 5\pm \sqrt{33} }{ 2 }\] \[\frac{ 5\pm \sqrt{17} }{ 2 }\] \[\frac{ *5\pm \sqrt{33} }{ 2 }\] \[\frac{ -5\pm \sqrt{17} }{ 2 }\]

OpenStudy (anonymous):

yes the x2 is x^2, and i just put more detailed answers to help

OpenStudy (anonymous):

@nfcfox

OpenStudy (nfcfox):

Kinda looks like they used the quadratic formula.

OpenStudy (nfcfox):

As it cannot be factored.

OpenStudy (nfcfox):

Let me evaluate.

OpenStudy (anonymous):

Okay

OpenStudy (anonymous):

and yes it deals with the quadratic formula

OpenStudy (nfcfox):

So let's add the 2. Our equation is now \[x ^{2} +5x+2=0\]

OpenStudy (anonymous):

Okay im following

OpenStudy (anonymous):

Now would we insert this into the quadratic formula?

OpenStudy (nfcfox):

The quadratic formula is: \[(b \pm \sqrt{b ^{2}-4ac})/2a\]

OpenStudy (nfcfox):

Where b is the middle term, a is the first, and c is the third.

OpenStudy (anonymous):

okay, and so i insert the data that we have into the formula correct?

OpenStudy (nfcfox):

Yes B: 5 A: 1 C: 2

OpenStudy (nfcfox):

Tell me if you figured it out.

OpenStudy (anonymous):

\[\frac{ 5\pm \sqrt{5^2-4(2)} }{ 2 }\]

OpenStudy (anonymous):

and then i simplify...

OpenStudy (nfcfox):

Yep.

OpenStudy (anonymous):

\[\frac{ 5\pm \sqrt{25-8} }{ 2}\] \[\frac{ 5\pm \sqrt{17} }{ 2 }\]

OpenStudy (anonymous):

which gives me B

OpenStudy (anonymous):

Thanks so much!!

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