Question

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`
....

Answer 1

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.

Answer 2

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`