How to solve A method for ax^3+bx^2+cx+d=0?

ax^3+bx^2+cx+d=0
atsqrt(t)+bt+csqrt(t)+d=0
(at+c)sqrt(t)+bt=-d
(at+c)^2t+2(at+c)sqrt(t)+b^2t^2=d^2
at+c=p, b^2=q, -d^2=r
p^2t+2psqrt(t)+qt^2+r=0
t=m/(p^2)
m+q/p^4m^2=-r-2sqrt(m)
-r-2sqrt(m)=-sqrt(r^2+4(rsqrt(m)+m)
q/p^4=n
m+nm^2=-sqrt(r^2+4(rsqrt(m)+m)
m^2+n^2m^4=-2nm^3-sqrt(r^2+4s
s^2=m^2+(r^2+2rsqrt(m))m
....

2 Answers
Jul 14, 2018

See explanation...

Explanation:

Given:

ax^3+bx^2+cx+d = 0" " with a != 0

We have:

0 = 27a^2(ax^3+bx^2+cx+d)

color(white)(0) = 27a^3x^3+27a^2bx^2+27a^2cx+27a^2d

color(white)(0) = (3ax)^3+3(3ax)^2b+3(3ax)b^2+b^3+(27a^2c-9ab^2)x+(27a^2d-b^3)

color(white)(0) = (3ax+b)^3+(9ac-3b^2)(3ax+b)+(2b^3+27a^2d-9abc)

color(white)(0) = t^3+pt+q

where:

{ (t = 3ax+b), (p = 9ac-3b^2), (q = 2b^3+27a^2d-9abc) :}

If the cubic has one real root and two non-real roots, then a good method from this point may be provided by Cardano's method:

Let t = u+v.

Then:

u^3+v^3+(3uv+p)(u+v)+q = 0

To eliminate the term in (u+v), add the constraint v = -p/(3u) to get:

u^3-p^3/(27u^3)+q = 0

Multiply through by 27u^3 to get the "quadratic in u^3":

27(u^3)^2+27q(u^3)-p^3 = 0

We can then use the quadratic formula to find:

u^3 = (-27q+-sqrt(729q^2+108p^3))/54

If this is real valued, then so is v^3 and by symmetry we can use + for u^3 and - for v^3 to find real root:

t_1 = root(3)((-27q+sqrt(729q^2+108p^3))/54)+root(3)((-27q-sqrt(729q^2+108p^3))/54)

color(white)(t_1) = 1/3 (root(3)((-27q+3sqrt(81q^2+12p^3))/2)+root(3)((-27q-3sqrt(81q^2+12p^3))/2))

and associated non-real complex roots:

t_2 = 1/3 (omega root(3)((-27q+3sqrt(81q^2+12p^3))/2)+omega^2 root(3)((-27q-3sqrt(81q^2+12p^3))/2))

t_3 = 1/3 (omega^2 root(3)((-27q+3sqrt(81q^2+12p^3))/2)+omega root(3)((-27q-3sqrt(81q^2+12p^3))/2))

where omega = -1/2+sqrt(3)/2i is the primitive complex cube root of 1.

Then we can use x = 1/(3a)(t - b) to find the corresponding roots of the original cubic.

If the given cubic has 3 real roots, then the quadratic in u^3 will have complex roots and we end up with slightly messier formulas based on v = -p/(3u), namely things like:

t = 1/3 root(3)((-27q+3sqrt(81q^2+12p^3))/2) - p/(root(3)((-27q+3sqrt(81q^2+12p^3))/2))

Note that in this expression, the cube roots will be irreducible complex cube roots, with a sum that is real. This is not a nice way of representing real roots. If the cubic has 3 real roots then it is probably better to avoid Cardano's method and use a trigonometric substitution instead.

Jul 15, 2018

Here's a method you can use in the case of 3 real roots...

Explanation:

Supplemental to the other answer, if the reduced cubic equation:

t^3+pt+q = 0

has 3 (irrational) real solutions, then my preference from that point is to use a trigonometric substitution:

t = k cos theta

Then we have:

k^3 cos^3 theta + k p cos theta + q = 0

In consideration of the trigonometric identity:

cos 3 theta = 4 cos^3 theta - 3 cos theta

choose:

k = 2sqrt(-p/3)

Then:

0 = k^3 cos^3 theta + k p cos theta + q

color(white)(0) = - 2/3 p sqrt(-p/3) (4 cos^3 theta - 3cos theta) + q

color(white)(0) = - 2/3 p sqrt(-p/3) cos 3 theta + q

and hence:

cos 3 theta = (3q)/(2 p sqrt(-p/3))

So:

3 theta = +-cos^(-1)((3q)/(2 p sqrt(-p/3)))+2npi" " for any integer n

So:

theta = +-1/3cos^(-1)((3q)/(2 p sqrt(-p/3)))+(2npi)/3

Remembering that cos (-theta) = cos (theta) and cos has period 2pi, this will give distinct values of cos theta:

cos theta = cos(1/3cos^(-1)((3q)/(2 p sqrt(-p/3)))+(2npi)/3)" " for n = 0, 1, 2

and hence three solutions of the given reduced cubic equation:

t_n = 2sqrt(-p/3) cos(1/3cos^(-1)((3q)/(2 p sqrt(-p/3)))+(2npi)/3)" " for n = 0, 1, 2