# Find real answers p and q

#### anemone

##### MHB POTW Director
Staff member
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

##### MHB POTW Director
Staff member
Solution suggested by other:

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 $q=12$.

#### Albert

##### Well-known member
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:

$6x^2-24x-4p=0---(1)$
$x^3+px^2+qx-8=0---(2)$
$(2)\times 6:6x^3+6px^2+6qx-48=0---(3)$
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 :
$p=-6, \,\, q=12\,\, and\,\, k=2$

Last edited:

#### anemone

##### MHB POTW Director
Staff member
my solution:

$6x^2-24x-4p=0---(1)$
$x^3+px^2+qx-8=0---(2)$
$(2)\times 6:6x^3+6px^2+6qx-48=0---(3)$
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 $q\ne 11$.