MHB POTW Director
- Feb 14, 2012
The equation $x^3+ax+b=0$ has three distinct real roots. Show that $a<0$.
Well done, MarkFL! I solved this problem using an entirely different approach, and hence you have just given me another insight about how to tackle it using the way you did... for this I want to say thank you!My solution:
We see that if $a=0$, then if $b=0$ the roots are not distinct. By Descartes' rule of signs, if $b>0$, there will only be one negative real root and if $b<0$ there will only be one positive real root. So, $a\ne0$.
Next, let's consider if $b=0$, then we have:
\(\displaystyle x\left(x^2+a \right)=0\)
We see that we must have $a<0$ in order to have 3 distinct real roots.
Next, let's consider if $b>0$.
By Descartes' rule of signs, and given that there are 3 distinct real roots, there will be 2 positive roots if there are 2 sign changes between successive coefficients. That is if $a<0$. Mutiplying the odd-powered terms by -1, we then have one sign change and thus there will be 1 negative root.
And finally, let's consider if $b<0$.
Descartes' rule of signs tells us that if $a$ is positive then there can only be at most 1 positive root. If $a$ is negative then there will be 1 positive root and 2 negative roots.
Thus, in all cases concerning $b$ we find that $a<0$.
My solution uses calculus.The equation $x^3+ax+b=0$ has three distinct real roots. Show that $a<0$.
Thanks, Opalg for participating! Your calculus approach seems so nice and neat!My solution uses calculus.
If $f(x) = x^3+ax+b$ then $f'(x) = 3x^2 + a$.
If $a>0$ then $f'(x)$ can never be zero, so $f$ has no turning points and can therefore cross the real axis only once. Thus $f(x) = 0$ has only one real solution.
If $a=0$ then the equation $f(x) = 0$ becomes $x^3 + b = 0$, whose only real root is the cube root of $-b$.
The only remaining possibility is that $a<0$, so that condition is necessary for the equation to have three real roots.