andriod09
  • andriod09
Need help on Quadratic equations please! The equation is on the comments.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
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andriod09
  • andriod09
\[4x^{2}+9x=0\] i know that the answer is: \[x=\frac{ -9\pm \sqrt{81-4(4)(2)} }{ 8 }\]
anonymous
  • anonymous
Much easier to solve your problem would be to factor out an x,\[x(4x+9)=0\]now you have two solutions,\[x=0\]and\[4x+9=0\]\[4x=-9\]\[x=-\frac{9}{4}\]and you get
anonymous
  • anonymous
The equation is correct. But you put innocent "2" instead of "0" for c.

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andriod09
  • andriod09
huh? the thing messed up, give me a sec
anonymous
  • anonymous
Well, then you can do the following:\[x_{1/2}=\frac{-9 \pm \sqrt{9^2-4*4*0}}{2*4}\]simplifying that you get,\[x_{1/2}=\frac{ -9 \pm 9 }{ 8 }\]Compute the rest
andriod09
  • andriod09
I have to use: \[ax^{2}+bx+c=0\] and i have to use: \[x=\frac{ -b\pm \sqrt{b^{2}-4ac} }{ 2a }\]
anonymous
  • anonymous
Yes.. that is what I have used. Just make c=0, and you will get the same result as I have.
andriod09
  • andriod09
but c = 2... thats what the answer says.
anonymous
  • anonymous
Well then you have given us a wrong problem\[4x^2+9x=0\]does not have a c in it..
andriod09
  • andriod09
my bad... /facepalm its: \[4x^{2}+9x+2=0\]
andriod09
  • andriod09
@gezimbasha
anonymous
  • anonymous
Hehe, no worries. Now you have to use the formula you input\[x_{1/2}=\frac{-9 \pm \sqrt{9^2-4*4*2}}{2*4}\]compute the values inside the square root first\[x_{1/2}=\frac{-9 \pm \sqrt{81- 32}}{8}\]and complete calculation\[x_{1/2}=\frac{-9 \pm 7}{8}\] and you get,\[x_1= -2\]\[x_2=-\frac{1}{4}\]
andriod09
  • andriod09
where did you get \[x _{1/2} \]
anonymous
  • anonymous
That just means that there are two solutions we will get from the equation. Its just mathematical notation. I am used to writing it like that. Don't worry, you can write just x, it will be fine.
andriod09
  • andriod09
im only in high school. :/ :L
anonymous
  • anonymous
Yeah I figured. The quadratic equations are studied extensively in high school. You're doing good tho. Might want to reconsider your profile picture heh ;)
andriod09
  • andriod09
it says i don't hate math. open my profile and look at the white! everyone says that. lol do you not see the white scribble on the \[^{don't}\] I hate math
anonymous
  • anonymous
Oh.. then I owe you an apology. XD.
andriod09
  • andriod09
Its okay. I need to make it bigger anyway. lol. Thanks tho.
anonymous
  • anonymous
You're welcome man. Anytime!
andriod09
  • andriod09
I gave you a medal. dont for get to give one back~ ;)
anonymous
  • anonymous
Hey, my bad, my first day over here.
andriod09
  • andriod09
what do you mean?
anonymous
  • anonymous
This is my first day in open study, so I'm still getting used to the medals and stuff.
andriod09
  • andriod09
how would i set up: \[3x^{2}+11x+5=0\] and oh. i didn't know
anonymous
  • anonymous
Same principle, replace a=3, b=11, and c=5.
andriod09
  • andriod09
i mean in the fraction form

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