# Prove: we always have at least x>0 is a $x^3+bx^2+cx-d^2=0$ 's root

Prove: we always have at least one x>0 is a $x^3+bx^2+cx-d^2=0$ 's root (b, c, d are real numbers and $d≠0$)

-
– user31280 Nov 10 '12 at 13:19
I think you need $d \neq 0$ for strict inequality. What happens at $x=0$ and $x \to \infty$? – WimC Nov 10 '12 at 13:21
Yes $d≠0$, sorry – Xeing Nov 10 '12 at 13:34
Use Descartes' rule of signs. – i. m. soloveichik Nov 10 '12 at 13:41

## 3 Answers

The statement you mentioned does not hold when d is allowed to be zero. Consider $x^3$ as a counterexample.I will assume that d is different from zeroLet p(x) be the polynomial.thus p(0)<0. Since p(x) tends to infinity as x goes to infinity. Therefore there exists r>0 such that p(r)>0. Now apply the intermediate value theorem on the interval [0,r]

-

Let, $$p(x) = x^3 + bx^2 + cx - d^2$$ Then we have, $p(0)= -d^2 < 0$ We know that there exists $x > 0$ so that: $$x^3 > bx^2 + cx - d^2$$ For any set of constants $b, c, d$. Let's call this pont $r$. Thus, we know that $p(0) < 0$ and $p(r)>0$ and that $r > 0$.

You can continue the rest.

-

If $f(x)=x^3+bx^2+cx-d^2$,

using this, there is only one change of sign in $f(x),$ so $f(x)$ can have at most one positive real root and the number of positive real root will be exactly one as the number of positive roots of the polynomial=the number of sign changes$(m)-2\cdot n$, where $0\le 2n\le m$.

-