MHB Prove Real Roots of $x^3+ax+b=0$ When $a<0$

  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Roots
AI Thread Summary
The equation $x^3 + ax + b = 0$ has three distinct real roots only when $a < 0$. If $a > 0$, the derivative $f'(x) = 3x^2 + a$ does not equal zero, indicating no turning points and only one real solution. When $a = 0$, the equation simplifies to $x^3 + b = 0$, which also yields a single real root. Therefore, the condition $a < 0$ is necessary for the existence of three distinct real roots. This conclusion is supported by calculus and the behavior of the function's derivative.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
The equation $x^3+ax+b=0$ has three distinct real roots. Show that $a<0$.
 
Mathematics news on Phys.org
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:

$$x\left(x^2+a \right)=0$$

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 $a<0$.
 
MarkFL said:
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:

$$x\left(x^2+a \right)=0$$

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 $a<0$.

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!:)
 
anemone said:
The equation $x^3+ax+b=0$ has three distinct real roots. Show that $a<0$.
My solution uses calculus.
[sp]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.[/sp]
 
Opalg said:
My solution uses calculus.
[sp]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.[/sp]

Thanks, Opalg for participating!:) Your calculus approach seems so nice and neat!:o

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.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top