need help one more time on solving quadratic equation

- anonymous

need help one more time on solving quadratic equation

- jamiebookeater

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- anonymous

ok

- anonymous

\[y ^{2}=-24y+144\] and so far I have \[-3 + or -\sqrt{57}\over -12\]

- anonymous

I'm sure it's supposed to simplify from there but not sure exactly how. Or if this will give me 2 non real answers, or 2 real ones

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## More answers

- anonymous

3 and 19 make 57

- anonymous

\[-24\pm \sqrt{b ^{}}\]

- anonymous

I got 0 under radical so -24+ -0/2

- anonymous

-12

- anonymous

ok, so -24 + and - -12 are the non real answers?

- anonymous

wait! I just realized I typed the wrong problem..

- anonymous

I used +144 not -144 sorry my oversight

- anonymous

Sorryy, can you guys wait a sec while I put the right one up here

- anonymous

Mary, when you're solving these, be sure to get them in the form
\[ax^2 + bx + c = 0\] before plugging into the quadratic equation

- anonymous

\[-6^{2}+3x+2=0\]

- anonymous

So what is a, b, and c?

- anonymous

b is 3 and c is 2

- anonymous

I'm guessing that first term should be \(-6x^2\)

- anonymous

so a is?

- anonymous

oh yeah, sorry i thought you just asked about b and c

- anonymous

a is -6

- anonymous

Ok, now what do you get when you plug them in?

- anonymous

the formula we have been going off of is a bit different than that. But I had gotten -3 + and - root 57 over -12

- anonymous

\[x=\frac{-3 \pm \sqrt{9-4(-6)(2)}}{2(-6)}\]
\[x = \frac{-3 \pm \sqrt{9+48}}{-12}\]
\[x = \frac{3 \pm \sqrt{57}}{12}\]

- anonymous

81 under radical bc of double neg so you add

- anonymous

sorry ya'll numbers all wrong I will leave

- anonymous

don't forget -3

- anonymous

ok so I had the right answer, except it's positive 3, not negative?

- anonymous

are they real or non real solutions?

- anonymous

You can simplify a bit further, but it's not gonna be a whole number or even a rational one.
\[x = \frac{3 \pm \sqrt{57}}{12}\]
\[x = \frac{3}{12} \pm \frac{\sqrt{57}}{12}\]
\[x = \frac{1}{4} \pm \frac{\sqrt{57}}{12}\]

- anonymous

They are real solutions. You only get non-real solutions when the thing under the square root is negative.

- anonymous

As for the -3, if you factor a -1 from the top and bottom you can cancel to get a positive 3 on top and a 12 on bottom. So it's going to be positive. It doesn't matter much though because one of your x will be positive and one is negative.

- anonymous

Your answer was also correct, but could be simplified a little bit.

- anonymous

ok, thanks!

- anonymous

because the - was common to both the numerator and denominator.

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