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anonymous
 5 years ago
need help one more time on solving quadratic equation
anonymous
 5 years ago
need help one more time on solving quadratic equation

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y ^{2}=24y+144\] and so far I have \[3 + or \sqrt{57}\over 12\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[24\pm \sqrt{b ^{}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got 0 under radical so 24+ 0/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, so 24 + and  12 are the non real answers?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait! I just realized I typed the wrong problem..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I used +144 not 144 sorry my oversight

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorryy, can you guys wait a sec while I put the right one up here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Mary, 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
 5 years ago
Best ResponseYou've already chosen the best response.0So what is a, b, and c?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm guessing that first term should be \(6x^2\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh yeah, sorry i thought you just asked about b and c

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, now what do you get when you plug them in?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the formula we have been going off of is a bit different than that. But I had gotten 3 + and  root 57 over 12

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x=\frac{3 \pm \sqrt{94(6)(2)}}{2(6)}\] \[x = \frac{3 \pm \sqrt{9+48}}{12}\] \[x = \frac{3 \pm \sqrt{57}}{12}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.081 under radical bc of double neg so you add

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry ya'll numbers all wrong I will leave

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so I had the right answer, except it's positive 3, not negative?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are they real or non real solutions?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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
 5 years ago
Best ResponseYou've already chosen the best response.0They are real solutions. You only get nonreal solutions when the thing under the square root is negative.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0As 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
 5 years ago
Best ResponseYou've already chosen the best response.0Your answer was also correct, but could be simplified a little bit.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because the  was common to both the numerator and denominator.
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