Can the Roots of a Cubic Equation be Bounded?

[anemone]

Let $p,\,q,\,r$ be real numbers such that the roots of the cubic equation $x^3+px^2+qx+r=0$ are all real. Prove that these roots are bounded above by $\dfrac{2\sqrt{p^2-3q}-p}{3}$.

 

[Albert1]

Let $p,\,q,\,r$ be real numbers such that the roots of the cubic equation $x^3+px^2+qx+r=0$ are all real. Prove that these roots are bounded above by $\dfrac{2\sqrt{p^2-3q}-p}{3}$.

Spoiler

let:$f(x)=x^3+px^2+qx+r---(1)=(x-a)(x-b)(x-c)$
here $a,b,c\in R$
and :$a\geq b\geq c$
suppose $f(x)$ degenerates to $(x-a)^3=x^3-3ax^2+3a^2x-a^3---(2)$
compare(1)(2)$p=-3a,\, q=3a^2,\,\, r=-a^3$
$a$ must be the inflection point of $f(x)$
using the first and second derivative test
it is clear that :
these roots are bounded above by $\dfrac{2\sqrt{p^2-3q}-p}{3}$
that is :$a\leq \dfrac{2\sqrt{p^2-3q}-p}{3}$

 

[Opalg]

Let $p,\,q,\,r$ be real numbers such that the roots of the cubic equation $x^3+px^2+qx+r=0$ are all real. Prove that these roots are bounded above by $\dfrac{2\sqrt{p^2-3q}-p}{3}$.

Proof by sketch:
[sp]

View attachment 3834​

For the cubic to have three real roots, the $x$-axis must lie between the two horizontal lines in the sketch. The largest root must therefore be less than (the $x$-coordinate of) the point marked X in the sketch, where the graph crosses the red line.

If the constant term in the polynomial is chosen so that the red line is the $x$-axis, then the largest root will occur at X, and the other two roots will be at Y, where the polynomial has its lower critical point. The critical points are given by the roots of the derivative of the polynomial, which is $3x^2 + 2px + q$. So the lower root is $\frac13\bigl(-p - \sqrt{p^2 - 3q}\bigr).$

The sum of the three roots of the original polynomial is independent of the constant term, and is always equal to $-p$. Therefore $\frac23\bigl(-p - \sqrt{p^2 - 3q}\bigr) + \alpha = -p$, where $\alpha$ is the $x$-coordinate of the point X. Thus $\alpha = \frac13\bigl(-p + 2\sqrt{p^2 - 3q}\bigr)$, as required.[/sp]

 

[Albert1]

Proof by sketch:
[sp]

View attachment 3834​

For the cubic to have three real roots, the $x$-axis must lie between the two horizontal lines in the sketch. The largest root must therefore be less than (the $x$-coordinate of) the point marked X in the sketch, where the graph crosses the red line.

If the constant term in the polynomial is chosen so that the red line is the $x$-axis, then the largest root will occur at X, and the other two roots will be at Y, where the polynomial has its lower critical point. The critical points are given by the roots of the derivative of the polynomial, which is $3x^2 + 2px + q$. So the lower root is $\frac13\bigl(-p - \sqrt{p^2 - 3q}\bigr).$

The sum of the three roots of the original polynomial is independent of the constant term, and is always equal to $-p$. Therefore $\frac23\bigl(-p - \sqrt{p^2 - 3q}\bigr) + \alpha = -p$, where $\alpha$ is the $x$-coordinate of the point X. Thus $\alpha = \frac13\bigl(-p + 2\sqrt{p^2 - 3q}\bigr)$, as required.[/sp]

according to Opalg's sketch:

Spoiler

$f(x)=x^3+px^2+qx+r=(x-a)(x-b)^2$
where $a>b,$ and $a,b\in R$
the coordinates of $x,y $ being :$x(a,0),y(b,0)$
and point $y$ happens to be relative maximum or relative minimum point of $f(x)$

 

[anemone]

Spoiler

let:$f(x)=x^3+px^2+qx+r---(1)=(x-a)(x-b)(x-c)$
here $a,b,c\in R$
and :$a\geq b\geq c$
suppose $f(x)$ degenerates to $(x-a)^3=x^3-3ax^2+3a^2x-a^3---(2)$
compare(1)(2)$p=-3a,\, q=3a^2,\,\, r=-a^3$
$a$ must be the inflection point of $f(x)$
using the first and second derivative test
it is clear that :
these roots are bounded above by $\dfrac{2\sqrt{p^2-3q}-p}{3}$
that is :$a\leq \dfrac{2\sqrt{p^2-3q}-p}{3}$

Thanks for participating and your solution, Albert.:)

Proof by sketch:
[sp]

View attachment 3834​

For the cubic to have three real roots, the $x$-axis must lie between the two horizontal lines in the sketch. The largest root must therefore be less than (the $x$-coordinate of) the point marked X in the sketch, where the graph crosses the red line.

If the constant term in the polynomial is chosen so that the red line is the $x$-axis, then the largest root will occur at X, and the other two roots will be at Y, where the polynomial has its lower critical point. The critical points are given by the roots of the derivative of the polynomial, which is $3x^2 + 2px + q$. So the lower root is $\frac13\bigl(-p - \sqrt{p^2 - 3q}\bigr).$

The sum of the three roots of the original polynomial is independent of the constant term, and is always equal to $-p$. Therefore $\frac23\bigl(-p - \sqrt{p^2 - 3q}\bigr) + \alpha = -p$, where $\alpha$ is the $x$-coordinate of the point X. Thus $\alpha = \frac13\bigl(-p + 2\sqrt{p^2 - 3q}\bigr)$, as required.[/sp]

Aww...Opalg! This is the first time I have seen a graph drawn "manually" by you and I will treasure it forever! https://ci5.googleusercontent.com/proxy/Gp4I_WcAFzTXd1fWEYApnKN7VDI4w7PZVf6stZodczWdwX-qkfXJm2iGqZ2AnfEHwnPzHMQDRCcM5yUFhPsAyJTuDENZDlgdAaRMTXmf=s0-d-e1-ft#http://www.mathhelpboards.com/images/smilies/redface.png

Thank you so much for your neat solution, and very well done, Opalg!

There is another solution that I wanted to share too:

Spoiler

Let $a,\,b,\,c$ be the three real roots of $x^3+px^2+qx+r=0$ such that $a \ge b \ge c$.

Then, we see that we have

$a^3+pa^2+qa+r=0\tag{1}$.

Now, we factor the given cubic polynomial, knowing $x=a$ is a real root and also bearing in mind we have the newly found equation (1):

\(\displaystyle \begin{array}{c|rr}& 1 & p & q & r \\ a & & a & a^2+ap & a^3+a^2p+aq \\ \hline & 1 & a+p & a^2+ap+q & (a^3+a^2p+aq)+r=(-r)+r=0 \end{array}\)

This gives

$x^3+px^2+qx+r=(x-a)(x^2+(a+p)x+(a^2+ap+q))$

The quadratic polynomial $x^2+(a+p)x+(a^2+ap+q)$ has $b$ and $c$ as its two real roots, and hence its discriminant is non-negaitve, that is,

$(a+p)^2-4(1)(a^2+ap+q)\ge 0$

Solving the above inequality for $a$ we obtain

$a\le \dfrac{2\sqrt{p^2-3q}-p}{3}$, which completes the proof.