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Quadratic Equation - Simple

  1. Apr 1, 2008 #1
    I know this is a simple question but for some reason im getting stumped:

    Question: x-3=-x^2
    atempt: (x^2)+x = 3

    x^2 + x + 1/4= 3(4) + 1/4

    (x+1/2)^2= 3

    root everthing

    x+1/2= root 3
    x=+,- root 3 -1/2
     
  2. jcsd
  3. Apr 1, 2008 #2

    symbolipoint

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    Your initial algebraic steps were wrong. From the start, you should obtain:

    EDIT: bad information was HERE.

    from which the solutions are very plain (what?).
     
    Last edited: Apr 1, 2008
  4. Apr 1, 2008 #3
    -1x2 = 3???
     
  5. Apr 1, 2008 #4

    symbolipoint

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    Trying again: Your first steps were wrong. You should first obtain

    [tex] \[
    x^2 + x - 3 = 0
    \]
    [/tex]

    and then you can use general solution to quadratic equation OR complete square.
     
  6. Apr 1, 2008 #5
    Well, i don't really understand what u tried to do.

    [tex] x-3=-x^{2}=> x^{2}+x-3=0[/tex] Now do you could either try to factor this out, or apply directly the quadratic formula, do you know it?

    [tex]x_1_,_2=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/tex]

    In this particular case, i don't think it will factor nicely, so it is better to use the quadratic formula i provided u with.
    Can u go from here?
     
    Last edited: Apr 1, 2008
  7. Apr 1, 2008 #6
    sry im only in gr 11 and this was how i was taught... and that formula was never taught to me
     
  8. Apr 1, 2008 #7
    Well, you better learn it then, because not always will you be able to factor a quadratic eq. nicely.

    The general form of a quadratic equation is

    [tex] ax^{2}+bx+c=0[/tex] so now can you figure out what a,b and c are in your problem?
     
  9. Apr 1, 2008 #8
    oo i do know that thats the form i write my solution as, but for my question i stated above i dont know what im doing wrong to keep me from reaching that formula
     
  10. Apr 1, 2008 #9
    Aha i gotch ya!

    [tex]x^2 + x = 3=>x^{2}+2x\frac{1}{2}+\frac{1}{4}-\frac{1}{4}=3=>(x+\frac{1}{2})^{2}=3+\frac{1}{4}=>(x+\frac{1}{2})^{2}=\frac{13}{4}[/tex]

    You forgot to add that 1/4 to the 3 on your right hand side.
    Well, now you know what to do right?
     
  11. Apr 1, 2008 #10

    Integral

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    Where did the bolded factor of 4 come from?

    You should have.

    [tex] (x+ \frac 1 2)^2 = 3 \frac 1 4 [/tex]
     
  12. Apr 1, 2008 #11
    that was just multiplying the denominator by 4 so i could add fractions
     
  13. Apr 1, 2008 #12
    Well you should have multiplied then both the denominator and the numerator by 4.
     
  14. Apr 1, 2008 #13
    yea i know i just was too lazy to type it out >.<
     
  15. Apr 1, 2008 #14
    Well, that's why you got the wrong result then. Better not be lazy!
     
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