PLEASE help me solve this thing, im tried to the stress i need to know how to do this!! :(
\[2/5y ^{2}+1/5y=36/5\]
its a quadratic equation
Have you learned about the quadratic equation in your class yet?
yes, its the last part that kills me :(
\[y = (-b \pm \sqrt{b^2 - 4ac})/(2a)\]
In order to solve, use that formula. Get your equation into a standard setup by subtracting 36/5. This will leave you with: \[2/5y^2 + 1/5y - 36/5 = 0\] 2/5 = a 1/5 = b -36/5 = c
i try and get rid of fractions first, it bugs me
so it would be 2, 1, and 36
-36
Sometimes you just have to deal with it, math is annoying like that sometimes ;) Keep in mind that the formula I gave you will give you two answers.
I know i need two :) its the last part with factoring that bothers me
I just wasn't sure if you knew or not :p. The equation works whether or not you factor. Sometimes, you just can't factor. If you take calculus, things get 10x crazier. I get lost in those problems sometimes. Just gotta roll with the punches :)
im horrid at this, so thats gonna really suck :(
If you stick around a minute, I can figure out if there's a way to factor. For these problems, I just learned to go straight to the quadratic formula.
ok, please hurry if you can, i have to get this homework done soon, i dont have a good internet signal at home
You can actually factor out (1/5) in this problem. That will leave you with: \[(1/5)( 2y^2 + 1x + 36 )=0\] multiply each side by 5 \[2y^2 + 1x + 36 = 0\] Although, I'm not completely certain this will give you the same answer. I'm pretty sure that by factoring out the denominator, you get a different answer
that should be -36, by the way. Not +36.
I got 4 and -4.5, which is altogether different from the original answer. I would personally go with your first answers, but it's up to you to make that decision.
thats not what it wants tho
\[2x^2+x-36=0\]\[(x-4)(2x+9)=0\]so:\[x=4\]or\[x=-\frac{9}{2}=-4.5\]
omg thank you!!! *cries*
:-)
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