MHB Can the Roots of a Cubic Equation be Bounded?

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The discussion focuses on proving that the roots of the cubic equation \(x^3 + px^2 + qx + r = 0\) are bounded above by \(\frac{2\sqrt{p^2 - 3q} - p}{3}\) when all roots are real. A sketch illustrates that for three real roots, the x-axis must lie between two horizontal lines, with the largest root located at a specific point where the graph intersects a defined line. The critical points of the polynomial, derived from its derivative, help identify the bounds of the roots. The sum of the roots is shown to be independent of the constant term, leading to the conclusion about the upper bound. Overall, the proof effectively combines graphical analysis and algebraic properties of cubic equations.
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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 $\dfrac{2\sqrt{p^2-3q}-p}{3}$.
 
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anemone said:
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 $\dfrac{2\sqrt{p^2-3q}-p}{3}$.
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 :$a\leq \dfrac{2\sqrt{p^2-3q}-p}{3}$
 
Last edited:
anemone said:
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]
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]
 

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Opalg said:
Proof by sketch:
[sp]
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:
$f(x)=x^3+px^2+qx+r=(x-a)(x-b)^2$
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 $f(x)$
 
Albert said:
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.:)

Opalg said:
Proof by sketch:
[sp]
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!:cool:

There is another solution that I wanted to share too:
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):

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

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.
 
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