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Mathematics 6 Online
OpenStudy (anonymous):

What are the approximate solutions of 2x^2 + 9x = 8 to the nearest hundredth? x ≈ 1.22 and x ≈ 3.28 x ≈ −0.76 and x ≈ 5.26 x ≈ −1.22 and x ≈ −3.28 x ≈ −5.26 and x ≈ 0.76

OpenStudy (imstuck):

you could either move the 8 over and use the quadratic formula, or you could leave it as is and complete the square to find the solutions. Which do you prefer?

OpenStudy (imstuck):

probably the quadratic formula is easier.

OpenStudy (anonymous):

I'm new to this so whichever is easiest.

OpenStudy (imstuck):

QF. Do you know it?

OpenStudy (anonymous):

Not a clue.

OpenStudy (imstuck):

ok, here it is:

OpenStudy (imstuck):

\[x=\frac{ -b \pm \sqrt{(b)^{2}-4ac} }{ 2a }\]

OpenStudy (imstuck):

Your polynomial is as follows:\[2x ^{2}+9x-8=0\]with a = 2, b = 9, and c = -8.

OpenStudy (imstuck):

So we sub in our values and it looks like this:\[x=\frac{ -9\pm \sqrt{(9)^{2}-4(2)(-8)} }{ 2(2) }\]

OpenStudy (imstuck):

Doing that math you will get this:\[x=\frac{ -9\pm \sqrt{81+64} }{4 }\]

OpenStudy (anonymous):

So regarding the previous response, is the typical equation setup ax+bx+c=d or what?

OpenStudy (imstuck):

And simplifying even further gives us:\[x=\frac{ -9\pm \sqrt{145} }{4 }\]

OpenStudy (imstuck):

In answer to your question, the standard form of a quadratic equation, which is what we have here, is\[ax ^{2}+bx+c=0\]

OpenStudy (imstuck):

Now you have two solutions for which to find answers:\[x=\frac{ -9+\sqrt{145} }{ 4 },x=\frac{ -9-\sqrt{145} }{ 4 }\]

OpenStudy (anonymous):

Okay. So how do I find the actual answer to the question?

OpenStudy (imstuck):

Use the square root button on your calculator to get the values rounded to the nearest hundredth. Like this: The square root of 145 is 12.04159. So do the math like this then:\[x=\frac{ -9+12.04159 }{ 4 },x=\frac{ -9-12.04159 }{4 }\]

OpenStudy (imstuck):

The first solution is x = .7603, and the second solution is x = -5.260

OpenStudy (imstuck):

The second choice down is your answer.

OpenStudy (anonymous):

Okay thanks! I have a couple more I might need help on so ill just tag you.

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