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- Feb 14, 2012

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The equation $x^3+ax+b=0$ has three distinct real roots. Show that $a<0$.

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- #1

- Feb 14, 2012

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The equation $x^3+ax+b=0$ has three distinct real roots. Show that $a<0$.

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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 $a<0$.

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Well done,

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 $a<0$.

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- Feb 7, 2012

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My solution uses calculus.The equation $x^3+ax+b=0$ has three distinct real roots. Show that $a<0$.

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.

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- Feb 14, 2012

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Thanks,My solution uses calculus.

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.

My solution:

Let $m,\,n,\,k$ be the three real distinct roots for $y=x^3+ax+b$.

Then Vieta's Formula tells us:

$m+n+k=0$, $mn+nk+mk=a$

Notice that

$(m+n+k)^2=m^2+n^2+k^2+2(mn+nk+mk)$---(1)

Hence, by substituting the above two into (1) we get

$0=m^2+n^2+k^2+2a$

This equation holds true iff $a<0$, and we're done.