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Mathematics 12 Online
OpenStudy (anonymous):

What are the exact solutions of x^2 = 4 − 7x? x = x equals negative 7 plus or minus the square root of thirty-three all over 2 x = x equals negative 7 plus or minus the square root of sixty-five all over 2 x = x equals 7 plus or minus the square root of sixty-five all over 2 x = x equals 7 plus or minus the square root of thirty-three all over 2

OpenStudy (anonymous):

@IMStuck

OpenStudy (imstuck):

Setting it up in standard form we have\[x ^{2}+7x-4=0\]

OpenStudy (imstuck):

Then we factor this one just like the last one. You know, when it asks you to find the solutions of a quadratic, what it is asking you to do is find where x = 0, which ARE the solutions of a quadratic equation, or in other words, where the graph goes through the x axis, if it does. So we continue on...

OpenStudy (imstuck):

\[x=\frac{ -7\pm \sqrt{(-7)^{2}-4(1)(-4)} }{ 2(1) }\]

OpenStudy (imstuck):

This simplifies to:\[x=\frac{ -7\pm \sqrt{49+16} }{2 }\]which further simplifies to\[x=\frac{ -7\pm \sqrt{65} }{ 2 }\]

OpenStudy (imstuck):

So these are your two solutions, still in radical form:\[x=\frac{ -7+ \sqrt{65} }{ 2 },x=\frac{ -7-\sqrt{65} }{ 2 }\]

OpenStudy (imstuck):

The square root of 65 = 8.06223 So here are our solutions and then we will round them:\[x=\frac{ -7+8.06223 }{ 2 },x=\frac{ -7-8.06223 }{ 2 }\]

OpenStudy (anonymous):

So just positive and negative versions of the same answer.

OpenStudy (imstuck):

Doing that math you get x= .531 and x= -7.531

OpenStudy (imstuck):

Yes just positive and negatives. But as you can see, the answer are VERY different from one another!

OpenStudy (imstuck):

your answer is the second choice down. They left the square root of 65 in the answer and used the plus/minus instead of solving it for both values.

OpenStudy (imstuck):

Got this? I'm going on a bike ride with my son for a bit, but I'll be back soon; tag me if you need me, ok?

OpenStudy (imstuck):

And as always and ever, TY for the medal! You just raised my SmartScore a point!

OpenStudy (anonymous):

Thank you so much! And no problem. Lol

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