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andriod09

  • 3 years ago

Need help on Quadratic equations please! The equation is on the comments.

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  1. andriod09
    • 3 years ago
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    \[4x^{2}+9x=0\] i know that the answer is: \[x=\frac{ -9\pm \sqrt{81-4(4)(2)} }{ 8 }\]

  2. gezimbasha
    • 3 years ago
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    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

  3. imron07
    • 3 years ago
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    The equation is correct. But you put innocent "2" instead of "0" for c.

  4. andriod09
    • 3 years ago
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    huh? the thing messed up, give me a sec

  5. gezimbasha
    • 3 years ago
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    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

  6. andriod09
    • 3 years ago
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    I have to use: \[ax^{2}+bx+c=0\] and i have to use: \[x=\frac{ -b\pm \sqrt{b^{2}-4ac} }{ 2a }\]

  7. gezimbasha
    • 3 years ago
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    Yes.. that is what I have used. Just make c=0, and you will get the same result as I have.

  8. andriod09
    • 3 years ago
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    but c = 2... thats what the answer says.

  9. gezimbasha
    • 3 years ago
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    Well then you have given us a wrong problem\[4x^2+9x=0\]does not have a c in it..

  10. andriod09
    • 3 years ago
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    my bad... /facepalm its: \[4x^{2}+9x+2=0\]

  11. andriod09
    • 3 years ago
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    @gezimbasha

  12. gezimbasha
    • 3 years ago
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    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,\[x_1= -2\]\[x_2=-\frac{1}{4}\]

  13. andriod09
    • 3 years ago
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    where did you get \[x _{1/2} \]

  14. gezimbasha
    • 3 years ago
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    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.

  15. andriod09
    • 3 years ago
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    im only in high school. :/ :L

  16. gezimbasha
    • 3 years ago
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    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 ;)

  17. andriod09
    • 3 years ago
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    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

  18. gezimbasha
    • 3 years ago
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    Oh.. then I owe you an apology. XD.

  19. andriod09
    • 3 years ago
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    Its okay. I need to make it bigger anyway. lol. Thanks tho.

  20. gezimbasha
    • 3 years ago
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    You're welcome man. Anytime!

  21. andriod09
    • 3 years ago
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    I gave you a medal. dont for get to give one back~ ;)

  22. gezimbasha
    • 3 years ago
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    Hey, my bad, my first day over here.

  23. andriod09
    • 3 years ago
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    what do you mean?

  24. gezimbasha
    • 3 years ago
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    This is my first day in open study, so I'm still getting used to the medals and stuff.

  25. andriod09
    • 3 years ago
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    how would i set up: \[3x^{2}+11x+5=0\] and oh. i didn't know

  26. gezimbasha
    • 3 years ago
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    Same principle, replace a=3, b=11, and c=5.

  27. andriod09
    • 3 years ago
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    i mean in the fraction form

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