need help one more time on solving quadratic equation
ok
\[y ^{2}=-24y+144\] and so far I have \[-3 + or -\sqrt{57}\over -12\]
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
3 and 19 make 57
\[-24\pm \sqrt{b ^{}}\]
I got 0 under radical so -24+ -0/2
-12
ok, so -24 + and - -12 are the non real answers?
wait! I just realized I typed the wrong problem..
I used +144 not -144 sorry my oversight
Sorryy, can you guys wait a sec while I put the right one up here
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
\[-6^{2}+3x+2=0\]
So what is a, b, and c?
b is 3 and c is 2
I'm guessing that first term should be \(-6x^2\)
so a is?
oh yeah, sorry i thought you just asked about b and c
a is -6
Ok, now what do you get when you plug them in?
the formula we have been going off of is a bit different than that. But I had gotten -3 + and - root 57 over -12
\[x=\frac{-3 \pm \sqrt{9-4(-6)(2)}}{2(-6)}\] \[x = \frac{-3 \pm \sqrt{9+48}}{-12}\] \[x = \frac{3 \pm \sqrt{57}}{12}\]
81 under radical bc of double neg so you add
sorry ya'll numbers all wrong I will leave
don't forget -3
ok so I had the right answer, except it's positive 3, not negative?
are they real or non real solutions?
You can simplify a bit further, but it's not gonna be a whole number or even a rational one. \[x = \frac{3 \pm \sqrt{57}}{12}\] \[x = \frac{3}{12} \pm \frac{\sqrt{57}}{12}\] \[x = \frac{1}{4} \pm \frac{\sqrt{57}}{12}\]
They are real solutions. You only get non-real solutions when the thing under the square root is negative.
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.
Your answer was also correct, but could be simplified a little bit.
ok, thanks!
because the - was common to both the numerator and denominator.
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