Simple math question - quadratics [Solved]

blue powder

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.

.kbf?

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.

Pi-r8

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.

blue powder

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...

GenericFan

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

blue powder

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!

Savant

blue powder wrote: »

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.