Finding necessary and sufficient conditions on $b$ and $c$ for $x^3-3b^2x+c=0$ to have three distinct real roots

Question

The task is to find necessary and sufficient condition on $b$ and $c$ for the equation $x^3-3b^2x+c=0$ to have three distinct real roots.

Are there any formulas (such as $x_1x_2=c/a$ and $x_1+x_2=-b/a$ for roots in $ax^2+bx+c=0$), but for equations of 3rd power?

Answer 1

What you're looking for are Vieta's Formulas. They can be derived for an $n$-degree polynomial in a fairly simple way:

Let the roots be $r_1,r_2,\cdots r_n$, and let the first term be $a$ (the polynomial we're doing this with is $P(x)$). Then we have

$$a(x-r_1)(x-r_2)\cdots(x-r_n) = P(x)$$

One can expand the LHS to get the formulas for each coefficient given the roots and $a$, called Vieta's Formulas.

For $n=2$, we have

$$a(x-r_1)(x-r_2) = ax^2+bx+c$$

$$x^2-(r_1+r_2)x+r_1r_2 = x^2+\frac{b}{a}x+\frac{c}{a}$$

Equating coefficients gives

$$r_1r_2 = \frac{c}{a},\ \ \ r_1+r_2 = -\frac{b}{a}.$$

Answer 2

Hint:

the discriminant of a cubic equation of the form $x^3+px+q=0$ is $\Delta= -4p^3-27q^2$ and the equation has three real distinct solutions iff $\Delta > 0$ .

Or you can use Vieta's formulas.

Answer 3

First, if your polynomial has three distinct real roots, then its derivative has two distinct real roots (deRolle's theorem). In your case, it translates to the polynomial $3x^2-3b^2$ having two distinct roots. In other words, we require $b\ne 0$.

Second, by studying the graph of the function $x^3-3b^2x+c$, you can see that if its local minimum (it is in $x_+=|b|$) yields a positive value, then we have only one real root; if it yields a strictly negative value, then we have $3$ roots; and if it yields zero, then we have $2$ roots. THerefore, the condition becomes $x_+^3-3b^2x_++c<0$, or $$c < 2 |b|^3.$$

Third, for similar reasons, the value in local maximum (situated in $x_-=-|b|$) must be positive, therefore, $c > -2|b|^3$.

In other words, our conditions can be written as $|c| < 2|b|^3$.

Answer 4

Consider this problem geometrically, as a question of where the straight line $y=3b^2x-c$ meets the basic cubic curve $y=x^3$. For three distinct meeting points, first the gradient $3b^2$ of the line must be nonzero, so $b\neq0$, and the line must not be too high (large negative $c$) or too low (large positive $c$). To determine the range of $c$, we must look at the borderline cases where the line touches the curve. this will be when the gradient of the curve, given by $3x^2-3b^2$, matches the gradient of the line: $3x^2-3b^2=3b^2$, namely at $x=\pm b\surd2$. At these points, the $y$ values are respectively $\pm2b^3\surd2$. Substitute these $(x,y)$ values into the equation of the line to get $\pm2b^3\surd2=\pm3b^3\surd2-c.$ Thus $c=\pm b^3\surd2$. For three distinct real roots, therefore, $c$ must be strictly within this range: $-|b|^3\surd2<c<|b|^3\surd2,$ which may be written succinctly as $$c^2<2b^6.$$