Show that the cubic equation has one real roots

Question

Show that $x^3+ax+b=0$ has

a) only one real root when $a>0$

b) at most only one of it's roots are in $(-\sqrt{-a/3},\sqrt{-a/3})$ when $a<0$.

For a) I supposed that it had two real roots and i came to the conclusion that the derivative function never attains the value $0$, i don't know what to do in b). Can someone help with the editing, I am new in this site and I think that it has to do again with Rolle's Theorem.

Answer 1

For (a), if $a>0$, then $p(x)=x^3+ax+b$ is strictly increasing on $\Bbb R$ as $p^\prime(x)=3x^2+a>0$. Since $p$ is a cubic (odd degree) polynomial, so it has a real root. Also since $p$ is strictly increasing and $p(x)\to -\infty$ as $x\to-\infty$ and $p(x)\to \infty$ as $x\to \infty$, it follows that $p$ can not have more than one real root. (alternatively also you can argue by Rolle's theorem as if there are two distinct real roots of $p$, then we would have $3c^2+a=p^\prime(c)=0$, for some real $c$, which is impossible.)

For (b), if $a<0$, then also $p$ has a real root. Moreover since $p^\prime(x)=3x^2+a$ vanishes at $x=\pm\sqrt{\frac{-a} {3}}$. Since we know that between any two consecutive roots of $p^\prime$, there is at most one root of $p$ (as if there are two roots say $\alpha$ and $\beta$ of $p$ between two consecutive roots $\alpha',\beta'$ of $p^\prime$, then by Rolle's theorem there exists $\gamma$ such that $p^\prime(\gamma)=0$ violating the consecutivness of $\alpha'$ and $\beta'$). So there is there is atmost one $c\in(-\sqrt{\frac{-a} {3}},\sqrt{\frac{-a} {3}})=J$ (say) such that $p(c)=0$ (observe that such unique $c$ exists if $p(-\sqrt{\frac{-a} {3}})p(\sqrt{\frac{-a} {3}})<0$). Thus in $J$, there is atmost one $c\in J$ such that $p(c)=0$ (by the similar argument given above).

Answer 2

Derive the polynomial by $x$ and look where the Zeros of $f'(x)$ are. Now compute the second derivative and plug in the Zeros of $f'(x)$ you have obtained.

Now it holds $f''(x)=6x$ and for a negative $x$ the 2nd derivative tells us that the extremum has a curvature in the direction: right; for positive $x$ the direction of curvature is: left. This means now, by Rolle's Theorem that on the negative extremum there are only values $f(x)$ near $x_{extremum}$ which are smaller than $f(x_{extremum})$ (Can be visualized by a sketch). The extremum for positive $x$ has two values of $f(x)$ greater than $f(x_{extremum,pos.})$ by Rolle's Theorem. There will be only one value that is lying between These two extrema of $f'(x)$; this value will be the root.