anonymous
  • anonymous
Need help real quick, Its Quadratic Formula, I solved the question, but i wanna make sure my answer is right. The question is :
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
The answer I got was |dw:1441264767920:dw|
anonymous
  • anonymous
thank you! :) the answer on the worksheet says its |dw:1441265454395:dw| so could the worksheet be wrong?
skullpatrol
  • skullpatrol
Ooops :( Let's try again..

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anonymous
  • anonymous
okayy, i understand it, but i just dont know how it got that when i have a different answer lol.
anonymous
  • anonymous
so maybe i have to + and - it? idk
skullpatrol
  • skullpatrol
Try using your calculator and sub the numbers back in.
Jhannybean
  • Jhannybean
\[3x^2-2x=9\]\[x^2-\frac{2}{3}x=3\] \[c=\left(\frac{-\dfrac{2}{3}}{2}\right)=\left(-\frac{1}{3}\right)^2=\frac{1}{9}\] \[x^2-\frac{2}{3}x\color{red}{+\frac{1}{9}} = 3 \color{red}{+\frac{1}{9}}\]\[\left(x-\frac{1}{3}\right)^2= \frac{28}{9}\]\[x-\frac{1}{3} = \pm \frac{\sqrt{28}}{3}\]\[\boxed{x=\pm \frac{\sqrt{28}}{3}+\frac{1}{3}}\]
Jhannybean
  • Jhannybean
You can check your answer by completing the square, if you're familiar with that method.
Jhannybean
  • Jhannybean
And the answer can be simplified even further from what I have written btw.
anonymous
  • anonymous
This is the method we are supposed to use : \[x=-b \pm \sqrt{b^2-4ac}/2a\] over 2a
anonymous
  • anonymous
would it be the same
wolf1728
  • wolf1728
The first part of that formula is already divided by 2a
anonymous
  • anonymous
yeah
wolf1728
  • wolf1728
so what you typed would imply that you need to divide it by 2a again
Jhannybean
  • Jhannybean
Okay so put your formula in quadratic form, \(ax^2+bx+c=0 \longrightarrow 3x^2-2x-9=0\) identify your a, b and c and plug them in to the quadratic.
anonymous
  • anonymous
ohh okay isee
skullpatrol
  • skullpatrol
|dw:1441265969581:dw|
Jhannybean
  • Jhannybean
yay our answers match @skullpatrol :)
anonymous
  • anonymous
thanks! i see what i did wrong
skullpatrol
  • skullpatrol
thanks for asking :-)

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