Cubic equation

[Guest]

actually, not sure if it belongs into this group:

I have the following problem with the cubic equation:
20x^3+8^2+7x+6=0

Is there some kind of cubic equation (just as the quadratic one)? According to wikipedia there, is:

http://en.wikipedia.org/wiki/Cubic_function
(under the title: "critical points of a cubic function")

So I tried to apply it just like the quadratic function. Sadly, my root-funtion turns out to be negative:

the square root of: 8^2-3*20*7 --> negative, so not possible

what am I doing wrong??
would really appreciate if sbdy could help!!!

[Guest]

The section of wikipedia you are referring to calculates the turning points (also called stationary points) of the cubic, not the roots.
Because the determinant was negative there are no turning points for your equation, it just continually increases.

If you want to calculate the roots, you need to look at the section titled "general formula for roots".
http://en.wikipedia.org/wiki/Cubic_func ... _for_roots
Because there are no turning points there will be exactly one real root (and two complex roots).
Choose k=1 to get the real root. The formula you get at the end is horrible, you may want to use the below link instead.

https://www.wolframalpha.com/input/?i=20x^3%2B8x^2%2B7x%2B6%3D0
calculates the roots for you (click the "exact form" button in the "real solution" section).

Hope this helped,

R. Baber.

[HallsofIvy]

Yes, there is a "cubic formula" one form of it is given at
https://math.vanderbilt.edu/schectex/courses/cubic/

[Guest]

Here is my take on the "cubic formula":
For any a, b, [tex](a- b)^3= a^3- 3a^2b+ 3ab^2- b^3[/tex] and [tex]3ab(a- b)= 3a^2b- 3ab^2[/tex].

Adding those [tex](a- b)^2+ 3ab(a- b)= a^3- b^3[/tex].

Let x= a- b, p= 3ab, and [tex]q= a^3- b^3[/tex] so the equation become [tex]x^3+ px= q[/tex] or [tex]x^2+ px- q= 0[/tex]. That is a "reduced cubic"- "reduced" because there is no "[tex]x^2[/tex]". Given any cubic equation we can change the variable to remove the square term and give a "reduced" equation.

Now, given [tex]x^3+ px- q= 0[/tex], can we use [tex]p= 3ab[/tex] and [tex]q= a^3- b^3[/tex] to find a and b and so x= a- b?

Yes, we can! From [tex]p= 3ab[/tex], [tex]b= \frac{p}{3a}[/tex] so that [tex]a^3- b^3= a^2- \frac{p^3}{3^3a^3}= q[/tex]. Multiply both sides by [tex]a^3[/tex] to get [tex](a^3)^2- \frac{p^3}{3^3}= qa^3[/tex] or [tex](a^3)^2- q(a^3)- \left(\frac{p}{q}\right)^3= 0[/tex]. Let [tex]X= a^3[/tex] and that becomes the quadratic equation [tex]X^2- qX- \left(\frac{p}{q}\right)^3= 0[/tex].

By the quadratic formula, [tex]X= a^3= \frac{q\pm\sqrt{q^2- 4\left(\frac{p}{q}\right)^3}}{2}[/tex].

Then [tex]a= \sqrt[3]{\frac{q\pm\sqrt{q^2- 4\left(\frac{p}{q}\right)^3}}{2}}[/tex]
and [tex]b= \frac{p}{3a}[/tex].

We have the "[tex]\pm[/tex]" on the quadratic formula and may have three cube roots but some will be the same giving the possible three distinct roots for the cubic equation.

[Guest]

Very well explained!
Thank you.

[GeradHum]

You're close, but there's a small mistake in your equation:
You wrote 8², but maybe you meant 8x², otherwise it's not a cubic polynomial.

If the equation is:
20x³ + 8x² + 7x + 6 = 0,
then yes, you can solve it using the cubic formula, but it's quite complex.

A better approach is to try rational root theorem first (test small integer values like x = -1, -2...).

Let me know the exact form of the equation and I’ll help further!