# Approximate roots of a cubic

Let $a$, $b$, and $c$ be positive numbers. Consider the cubic equation $x^2(ax+b) = c$. Are there any useful bounds (upper and lower) I can put on the unique root of this equation for $x>0$? For example, we know that $x^* <(c/a)^{1/3}$ and $x^* < \sqrt{c/b}$. This equation can be solved explicitly, but that expression is rather complicated and I'm just looking to find an interval that contains a solution. Thanks!

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What's wrong with the bounds you have? – J.C. Ottem Mar 2 '11 at 18:34
Shouldn't the first one be $x^*<(c/a)^{1/3}$ - just lose the positive $x^2 b$ term? – Mark Bennet Mar 2 '11 at 20:29
Thanks Mark. Fixed that. – Jennifer Gao Mar 3 '11 at 4:28

## 1 Answer

Note that, since your $f(x):=x^2(ax+b)-c$ is convex on $[0,+\infty)$ the Newton's iteration with initial point $x_0>0$ produces a sequence $x_n$ which is decreasing for $n\ge1$ and converges to $x^*$. From the equation, $y_n:=c/ax_n^2\, -\, b/a$ also converges to $x^*$, but increasing. This gives you an interval $[y_n, x_n]$ whose endpoints are rational functions of the coefficients (and as small as you wish, for large enough $n$).

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The initial point for Newton's iteration should be $>0$, since $f'(0)=0$. – Julián Aguirre Mar 3 '11 at 9:06
right! fixed . – Pietro Majer Mar 3 '11 at 9:12