andriod09
  • andriod09
Okay people, I need help with Quadratic equations, i.e. \[ax^{2}+bx+c=0\] I have 5 and i will only put 1 here. so keep up!
Mathematics
katieb
  • katieb
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cgreenwade2000
  • cgreenwade2000
Sounds easy enough. DO you know the Quadratic formula?
andriod09
  • andriod09
\[4x^{2}+9x+2-0\] I know that the answer is: \[x=\frac{ -9\pm \sqrt{81-4(4)(2)} }{ 8 }\]
andriod09
  • andriod09
How do i get from point A to point B? aka from the equation to the answer

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andriod09
  • andriod09
yea, thats what it says in my book. I do the Life of Fred series and i am doing Quadratic Equation.
andriod09
  • andriod09
rly? its that easy?!?!?!?!?! omfg!
cgreenwade2000
  • cgreenwade2000
Take it a step further and come up with two different answers android. I believe in you.
andriod09
  • andriod09
how are ther two different answers?
andriod09
  • andriod09
Honestly, ill say it in the 3 different launguages i know. English:no Spanish:no ASL:shankes head
andriod09
  • andriod09
ik but, i don't get how you got them. how do you go from\[-9\pm \frac{ \sqrt{81-32} }{ 8 } \to -9\pm \frac{\sqrt{49} }{ 2 } \]
anonymous
  • anonymous
For \[ax^2+bx+c=0=4x^2+9x+2\] \[a=4,b=9,c=2\] using the quadratic formula \[x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\] substituting in \[a=4,b=9,c=2\] \[x=\frac{-9\pm \sqrt{9^2-4(4)(2)}}{2(4)}\] \[x=\frac{-9\pm \sqrt{81-32}}{8}\] \[x=\frac{-9 \pm \sqrt{49}}{8}\] \[x=\frac{-9\pm 7}{8}\] \[x=\frac{-9+7}{8}\] \[x=\frac{-2}{8}=\frac{-1}{4}\] and \[x=\frac{-9-7}{8}\] \[x=\frac{-16}{8}=-2\]
andriod09
  • andriod09
okay, delete that long post and spearate the posts by a \[\to\] that get put into each other so its easier to read.

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