pentic root

[Jon]

The root to ax^5+bx+c=0 is approximately,

x = { -c/({a^2+b^2}^(1/2)) }^(1/5)

Using this formula, the roots to,

x^5+x-34=0 x=1.888 should be 2
32x^5+4x-3=0 x=0.623 should be 1/2
x^5+x-0.10001 x=0.5887 should be 1/10

[Sam Wormley]

Post by Jon
The root to ax^5+bx+c=0 is approximately,
x = { -c/({a^2+b^2}^(1/2)) }^(1/5)
Using this formula, the roots to,
x^5+x-34=0 x=1.888 should be 2
32x^5+4x-3=0 x=0.623 should be 1/2
x^5+x-0.10001 x=0.5887 should be 1/10

http://www.wolframalpha.com/input/?i=roots+ax%5E5%2Bbx%2Bc%3D0

[hagman]

Post by Jon
The root to ax^5+bx+c=0 is approximately,
x = { -c/({a^2+b^2}^(1/2)) }^(1/5)
Using this formula, the roots to,
x^5+x-34=0  x=1.888  should be 2
32x^5+4x-3=0 x=0.623 should be 1/2
x^5+x-0.10001 x=0.5887 should be 1/10

That approximation is exact if either b=0 or c=0, but as your examples
show of doubtful quality in the general case

[Robert Israel]

The root to ax^5+bx+c=3D0 is approximately,
x =3D { -c/({a^2+b^2}^(1/2)) }^(1/5)
Using this formula, the roots to,
x^5+x-34=3D0 =A0x=3D1.888 =A0should be 2
32x^5+4x-3=3D0 x=3D0.623 should be 1/2
x^5+x-0.10001 x=3D0.5887 should be 1/10

That approximation is exact if either b=3D0 or c=3D0, but as your examples
show of doubtful quality in the general case

Here's a better approximation, for when b is small:

x = 1/a*(-c*a^4)^(1/5) + (-1/5*a^2/(-c*a^4)^(3/5))*b
+(-1/25*a^5/(-c*a^4)^(7/5))*b^2 + (-1/125*a^8/(-c*a^4)^(11/5))*b^3
+(-21/15625*a^2/(-c*a^4)^(4/5)/c^3)*b^5 + O(b^6)

On the other hand, when c is small, the root near 0 is approximately

(-1/b)*c + a/b^6*c^5 + (-5*a^2/b^11)*c^9 + O(c^13)