A cubic polynomial with real coefficients, $a x^3 + b x^2 + c x + d$, has either three real roots, or one real root and a pair of complex conjugate ones. If the latter happens, what is the explicit formula for this real solution, and what conditions can be placed on $a,b,c$ and $d$ to guarantee that the real root is positive?
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$\begingroup$ You are missing the assumption that it has real coefficients $\endgroup$– user208649Jan 17, 2016 at 18:21
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$\begingroup$ math.vanderbilt.edu/~schectex/courses/cubic may help $\endgroup$– Sean EnglishJan 17, 2016 at 18:29
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3 Answers
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For explicit formulas of the roots involving the coefficients, you may wish to consider Cardano's formula which will give it for degree 3 and 4.
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$\begingroup$ Thanks. A question: Wikipedia says if its discriminant is positive it has three distinct real roots, but in your link this is the case if the discriminant is negative? $\endgroup$– user305855Jan 17, 2016 at 18:28
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$\begingroup$ Most likely means their definitions of discriminant are off by a sign. $\endgroup$– FutureJan 17, 2016 at 18:38
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First, you need to eliminate the second degree term. For this, make $x=y+k$ with $k\in\mathbb{R}$, replace in the equation and determine $k$ in order to eliminate the second degree term.
Done that, we have an equation of the form $$ax^{3}+bx+c=0.$$ Writing $x=u+v$ and replace in the equation, you will find a quadratic equation whose roots are given in terms of $u$ and $v$ (in particular, of $x$).
Done that, it is easy to find the ultimate root.
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$\begingroup$ Very nice. Thanks. $\endgroup$– user305855Jan 17, 2016 at 18:37
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$\begingroup$ I am curious to know in which conditions we have two reals roots of opposite signs? $\endgroup$– ZbigniewJan 28, 2016 at 18:45
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$\begingroup$ @Zbigniew I do not know. Knows reciprocal polynomials? I'm not sure, but I think about then we are able to have two real roots of opposite signs. $\endgroup$– RafaelJan 29, 2016 at 0:16
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Case 1
If the derivative discriminant $ b^2 - 3 a c > 0 $
Let the roots of the derivative quadratic
$$ 3 a x^2 + 2 b x + c = 0 $$
be
$$ x_1,x_2 $$
and corresponding max/ min values be
$$ y_1, y_2 $$
If these above $y$ values have the same sign ( both positive or both negative or $ y_1 y_2 >0 $) then there is a single real root and two complex conjugates.
Draw a graph of these cubic and derivative functions to ascertain why it should be so.
If $ y_1, y_2 $ are of opposite sign or $ y_1 y_2 < 0 $ then there are three real roots,and,
if one of them is zero, two real roots among them are equal.
EDIT 1:
In terms of $ (a,b,c,d ), $
(If I made no error during algebraic simplification which please check,)
we have
$ \Delta > 0 $ for two imaginary complex roots & one real root, and, $ \Delta < 0 $ for three real roots, where
$$ 27 \Delta= $$
$$ (a c^2 ((-6 + (7 - 2 a) a) b^2 + c (-3 + a^2)^2 ) - 2 c ((-6 + 4 a) b^2 + 9 c (1 + a - a^2) ) d + 27 d^2). $$
Case 2
If the derivative discriminant $ b^2 - 3 a c < 0 $ then we have an inflection point on the x-axis.
answered Jan 17, 2016 at 18:56