Real Solutions for p and q in Quadratic and Cubic Equations

[anemone]

Find all pairs $(p,\,q)$ of real numbers such that the roots of quadratic and cubic equations $6x^2-24x-4p=0$, $x^3+px^2+qx-8=0$ are all non-negative real numbers.

 

[anemone]

Solution suggested by other:

Spoiler

Let $x_1,\,x_2$ be the roots of the first polynomial and $x_1',\,x_2',\,x_3'$ be the roots of the other polynomial.

By Viete's formulae, we have

1. $x_1+x_2=4$,

2. $x_1x_2=-\dfrac{2p}{3}$,

3. $x_1'+x_2'+x_3'=-p$,

4. $x_1'x_2'+x_2'x_3'+x_1'x_3'=q$,

5. $x_1'x_2'x_3'=8$

Note that $4=\left( \dfrac{4}{2} \right)^2=\left( \dfrac{x_1+x_2}{2} \right)^2 \ge x_1x_2=-\dfrac{2p}{3}$ and $-\dfrac{2p}{3}=\dfrac{2(x_1'+x_2'+x_3')}{3} \ge 2\sqrt[3]{x_1'x_2'x_3'}=4$.

We see that in both inequalities, equality actually holds. Consequently, we have

$x_1=x_2$, $x_1'=x_2'=x_3'$, and $-\dfrac{2p}{3}=4$

This gives $p=-6$ and $x_1'x_2'x_3'=8$ suggests $x_1'=x_2'=x_3'=2$ which gives $q=12$.

 

[Albert1]

Find all pairs $(p,\,q)$ of real numbers such that the roots of quadratic and cubic equations $6x^2-24x-4p=0$, $x^3+px^2+qx-8=0$ are all non-negative real numbers.

my solution:

Spoiler

$6x^2-24x-4p=0---(1)$
$x^3+px^2+qx-8=0---(2)$
$(2)\times 6:6x^3+6px^2+6qx-48=0---(3)$
we let $(3)=(1)\times (x-k),\,\,(here :\,k>0)$
if the roots of (1) are all non-negative real numbers
we can promise that the roots of (3) also are all non-negative real numbers
$\therefore 6x^3+6px^2+6qx-48=(6x^2-24x-4p)(x-k)$
expand the right side and compare with left side we have :
$6p=-6k-24---(4)$
$6q=24k-4p---(5)$
$4pk=-48---(6)$
from (4)(5)(6) we get :
$p=-6, \,\, q=12\,\, and\,\, k=2$

 

[anemone]

my solution:

Spoiler

$6x^2-24x-4p=0---(1)$
$x^3+px^2+qx-8=0---(2)$
$(2)\times 6:6x^3+6px^2+6qx-48=0---(3)$
we let $(3)=(1)\times (x-k),\,\,(here :\,k>0)$
if the roots of (1) are all non-negative real numbers
we can promise that the roots of (3) also are all non-negative real numbers
$\therefore 6x^3+6px^2+6qx-48=(6x^2-24x-4p)(x-k)$
expand the right side and compare with left side we have :
$6p=-6k-24---(4)$
$6q=24k-4p---(5)$
$4pk=-48---(6)$
from (4)(5)(6) we get :
$p=-6, \,\, q=12\,\, and\,\, k=2$

Thanks for participating, Albert!:)

But, I think you obtained the correct answer in your equating the two unrelated cubic equations was only a fluke...

Says if $(6x-1)(x-2)=6x^2-13x+2=6x^2-13x+p=0$ where $p=2$ and $(x-1)(x-2)(x-3)=x^3-6x^2+11x-6=x^3-3px^2+qx-6=0$ where $q=11$ are the two equations in question, then solving it your way resulted in $p\ne 2$ and $q\ne 11$.

 

[CellCoree]

I would approach this problem by first analyzing the given equations and understanding their behavior. The first equation is a quadratic equation with a leading coefficient of 6 and a constant term of -4p. The second equation is a cubic equation with a leading coefficient of 1 and a constant term of -8.

To find the solutions for p and q, we can use the discriminant and the Vieta's formulas. The discriminant of a quadratic equation is used to determine the nature of its roots. If the discriminant is greater than 0, the equation will have two distinct real roots. If the discriminant is equal to 0, the equation will have a repeated real root. And if the discriminant is less than 0, the equation will have two complex roots.

In this case, we want all the roots to be non-negative real numbers, which means that the discriminant of the first equation must be greater than or equal to 0. This can be written as:

$24^2-4(6)(-4p) \geq 0$

$576 + 96p \geq 0$

$p \geq -6$

Similarly, for the second equation, we can use the Vieta's formulas to find the sum and product of its roots. The sum of the roots is equal to -p and the product of the roots is equal to -8. Since we want all the roots to be non-negative, we can write the following inequality:

$p \geq 0$

Combining this inequality with the previous one, we get the final solution:

$p \geq 0$ and $p \geq -6$

Therefore, the range of values for p is $p \geq 0$.

To find the range of values for q, we can use the fact that the sum of the roots of a cubic equation is equal to -p and the product of the roots is equal to -8. This can be written as:

$q \geq -p$ and $q \geq 8$

Substituting the value of p from the previous inequality, we get:

$q \geq 0$ and $q \geq 8$

Therefore, the range of values for q is $q \geq 8$.

In conclusion, the real solutions for p and q in