Question

Condition that all roots of cubic equation are real

Answer 1

Suppose that (including multiplicity) the roots of

f(x)=Ax3+Bx2+Cx+D,

A≠0 are r1,r2,r3. Then, inspection shows that the quantity

D(f):=A4(r3−r2)2(r1−r3)2(r2−r1)2,

called the (polynomial) discriminant of f, vanishes iff f has a repeated root. (The coefficient A4 is unnecessary for the expression to enjoy this property, but among other things, its inclusion makes the below formula nicer.) On the other hand, with some work (say, by expanding and using Newton's Identities and Vieta's Formulas) we can write D(f) as a homogeneous quartic expression in the coefficients A,B,C,D:

D(f)=−27A2D2+18ABCD−4AC3−4B3D+B2C2.

It turns out that D gives us the finer information we want, too: f has three distinct, real roots iff D(f)>0 and one real root and two conjugate, nonreal roots iff D(f)<0.

It's apparent that one can generalize the notion of discriminant to polynomials p of any degree >1, producing an expression homogeneous of degree 2(degp−1) in the polynomial coefficients. In each case, up to a constant that depends on the degree and the leading coefficient of f, D(f) is equal to the resultant R(f,f′) of f and its derivative f′(x)=3Ax2+2Bx+C.

By making a suitable affine change of variables x⇝y, by the way, one can transform the given cubic to the form

f~(y)=y3+Py+Q

(which does not change the multiplicity of roots), and for a cubic polynomial in this form the discriminant has the simple and well-known form

−4P3−27Q2.

Answer 2

For a Cubic Equation :

$f(x) = a {x}^{3} + b {x}^{2} + cx + d = 0$

$\forall \: \: x_{1,2,3} \in \: ℝ$

The Condition For All Real Roots to Occur :

$\sf{b}^{2} {c}^{2} + 18abcd - 27 {a}^{2} {d}^{2} - 4a {c}^{3} - 4 {b}^{3} d > 0$

⇒ All the Roots of the Cubic Are :

$\sf \: x_{1,2,3} = \frac{ - b \: + \:2 \sqrt{ {b}^{2} - 3ac } \cos( \frac{1}{3} \arccos [ \frac{9abc - 2 {b}^{3} - 27 {a}^{2} d}{2( {b}^{2} - 3ac) \sqrt{ { {b}^{2} - 3ac} } } ] + \frac{2k\pi}{3} ) }{3a}$

where,

$k \in (0,1,2)$

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