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
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2 Answers
See explanation...
Explanation:
Given:
ax^3+bx^2+cx+d = 0" " witha != 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
Then:
u^3+v^3+(3uv+p)(u+v)+q = 0
To eliminate the term in
u^3-p^3/(27u^3)+q = 0
Multiply through by
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
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
Then we can use
If the given cubic has
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
Here's a method you can use in the case of
Explanation:
Supplemental to the other answer, if the reduced cubic equation:
t^3+pt+q = 0
has
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 integern
So:
theta = +-1/3cos^(-1)((3q)/(2 p sqrt(-p/3)))+(2npi)/3
Remembering that
cos theta = cos(1/3cos^(-1)((3q)/(2 p sqrt(-p/3)))+(2npi)/3)" " forn = 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)" " forn = 0, 1, 2