Simple math question - quadratics [Solved]

blue powderblue powder Registered User
edited November 2009 in Help / Advice Forum
I'm trying to refresh myself on math, and i've come across something I can't understand.

so we have the polynomial f(x)=4x^3+3x^2-6x-1=0, and the derivative, being a quadratic, is

f'(x)=12x^2+6x-6 and apparently this has explicit solutions.

so when I try and place these values into the quadratic equation, i got a different value to the text. they got -1 and 1/2. i soon realized that the text used b and b^2 from the polynomial, and the a and c values of the quadratic.

so what's the deal? i thought if you were using a quadratic equation, you only use the first derivative of the polynomial (the quadratic), not the polynomial. is it a process of using both the equations?

thanks in advance.

note: I'm using the text "mathemaical methods for physics and engineering" 2nd edition by k.f. riley, m.p. hobson and s.j. bence. and this is page 6.

blue powder on

Posts

  • .kbf?.kbf? Registered User
    edited November 2009
    What exactly is the question?

    If you are trying to find the zeros of f'(x) then yes, the answer is -1 and 1/2.
    12x^2+6x-6=0 -> x = (-6 +-[6^2-4(12)(-6)]^(1/2))/(2*12) = -1 and 1/2

    edit: just to clarify "f(x)=4x^3+3x^2-6x-1=0" doesn't make sense. You have a function f(x) which is 4x^3+3x^2-6x-1. At some points a function may be equal to zero. To find these points you use the equality 4x^3+3x^2-6x-1=0. f(x)=0 implies the function is always zero which is obviously not the case.

    .kbf? on
  • Pi-r8Pi-r8 Registered User regular
    edited November 2009
    I'm gonna go out on a limb and guess that you're looking for the min and max values of the original function. so you just plug -1 and 1/2 into it.

    Pi-r8 on
  • blue powderblue powder Registered User
    edited November 2009
    I'll go through those solutions when I have some free time, and I apologise for being so vague. The text actually used the quadratic equation, that being the -b+-(sqr root)(b^2-4ac)/(2a) kind of thing. but the text used values from both the first equation and the second equation. i thought you would just use the second equation, the derivative, to get the roots...

    blue powder on
  • GenericFanGenericFan Registered User
    edited November 2009
    You can solve the quadratic for 0 to get -1 and 1/2 using only values from the derivative:

    -6 +- sqrt(6^2 - (4*12*-6))/2*12

    -6 +- sqrt(36 - (-288))/24

    -6 +- 18 /24

    equals -24/24 and 12/24
    simplifies to -1 and 1/2

    GenericFan on
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  • blue powderblue powder Registered User
    edited November 2009
    ohh this is terribly embarrassing.. it seems that I simply wasn't puting the values in my calculator properly.

    Thank you all for your time and help!

    blue powder on
  • SavantSavant Registered User regular
    edited November 2009
    I'm trying to refresh myself on math, and i've come across something I can't understand.

    so we have the polynomial f(x)=4x^3+3x^2-6x-1=0, and the derivative, being a quadratic, is

    f'(x)=12x^2+6x-6 and apparently this has explicit solutions.

    so when I try and place these values into the quadratic equation, i got a different value to the text. they got -1 and 1/2. i soon realized that the text used b and b^2 from the polynomial, and the a and c values of the quadratic.

    so what's the deal? i thought if you were using a quadratic equation, you only use the first derivative of the polynomial (the quadratic), not the polynomial. is it a process of using both the equations?

    thanks in advance.

    note: I'm using the text "mathemaical methods for physics and engineering" 2nd edition by k.f. riley, m.p. hobson and s.j. bence. and this is page 6.

    Edit: nevermind, looks like you found out what was wrong.

    Savant on
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