MHB Evaluating $162p-172q$ from 3 Common Roots

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The equations given have three common roots, leading to the derived polynomial difference, which helps identify the fourth roots of each equation. The sum of the roots for the first equation is 1, while for the second it is -5, resulting in a fourth root of 3 for the first equation and -3 for the second. By substituting these roots back into the original equations, the values of p and q are determined to be 6 and -6, respectively. Consequently, the evaluation of 162p - 172q results in 2004. This calculation highlights the interconnectedness of the roots and coefficients in polynomial equations.
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The following equations have 3 common roots:

$x^4-x^3-7x^2+x+p=0$

$x^4+5x^3+5x^2-5x+q=0$

Evaluate $162p-172q$.
 
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Re: Evaluate 162p-172q

anemone said:
The following equations have 3 common roots:

$x^4-x^3-7x^2+x+p=0$

$x^4+5x^3+5x^2-5x+q=0$

Evaluate $162p-172q$.

Hello.

I have come to:

p=-q \rightarrow{}162p-172q=334p=-334q

If it is that, the solution sought, I show the test.:o

Regards.
 
Re: Evaluate 162p-172q

anemone said:
The following equations have 3 common roots:

$x^4-x^3-7x^2+x+p=0$

$x^4+5x^3+5x^2-5x+q=0$

Evaluate $162p-172q$.
[sp]The sum of the roots of the first equation is $1$. The sum of the roots of the second equation is $-5$. The three roots common to both equations will also be roots of the difference between the equations, namely $6x^3 + 12x^2 - 6x + q-p.$ The sum of the roots of that equation is $-2$. Therefore the fourth root of the first equation is $1-(-2) = 3$, and the fourth root of the second equation is $-5-(-2) = -3$. Putting $x=3$ in the first equation and $x=-3$ in the second equation, we get
$3^4 - 3^3 - 7\cdot3 + 3 + p = 0$, from which $p=6$,
$3^4 - 5\cdot3^3 + 5\cdot3 + 5\cdot3 + q = 0$, from which $q=-6$.​
Therefore $162p - 172q = 6(162+172) = 2004$ (presumably this problem was set in that year).[/sp]
 
Re: Evaluate 162p-172q

mente oscura said:
Hello.

I have come to:

p=-q \rightarrow{}162p-172q=334p=-334q

If it is that, the solution sought, I show the test.:o

Regards.

Hey mente oscura, yes, that is partially correct and you will gain full mark only if you have found the value of either $p$ or $q$ and substitute it into the equation.:)

Opalg said:
[sp]The sum of the roots of the first equation is $1$. The sum of the roots of the second equation is $-5$. The three roots common to both equations will also be roots of the difference between the equations, namely $6x^3 + 12x^2 - 6x + q-p.$ The sum of the roots of that equation is $-2$. Therefore the fourth root of the first equation is $1-(-2) = 3$, and the fourth root of the second equation is $-5-(-2) = -3$. Putting $x=3$ in the first equation and $x=-3$ in the second equation, we get
$3^4 - 3^3 - 7\cdot3 + 3 + p = 0$, from which $p=6$,
$3^4 - 5\cdot3^3 + 5\cdot3 + 5\cdot3 + q = 0$, from which $q=-6$.​
Therefore $162p - 172q = 6(162+172) = 2004$ (presumably this problem was set in that year).[/sp]

Awesome! Your approach is a great one and thanks for participating again, Opalg!

My solution:
I first let $a, b, c$ be the three common roots of those two equations and $m$ be the fourth root of the equation $x^4-x^3-7x^2+x+p=0$ and $n$ be the fourth root of the equation $x^4+5x^3+5x^2-5x+q=0$.

Thus, we can rewrite the given two equations in a different form, that is:

$x^4-x^3-7x^2+x+p=(x-m)(x^3-(a+b+c)x^2+(ab+bc+ca)x-abc)$-(1)

$x^4+5x^3+5x^2-5x+q=(x-n)(x^3-(a+b+c)x^2+(ab+bc+ca)x-abc)$-(2)

Divide the equations and cross multiplying gives

$(x-m)(x^4+5x^3+5x^2-5x+q)=(x-n)(x^4-x^3-7x^2+x+p)$

By comparing the coefficient of $x^4$ and $x^3$ from both sides of the equation, we get:

$5-m=-1-n$ hence $m-n=6$

$5-5m=-7+n$ and therefore $n+5m=12$

Solving the equations $m-n=6$ and $n+5m=12$ we get

$m=3$ and $n=-3$

Since $m$ and $n$ are one of the roots of the equations, if we substitute them back and let each of them equals zero, we will get the value of $p$ and $q$ respectively:

[TABLE="class: grid, width: 1000"]
[TR]
[TD]$x^4-x^3-7x^2+x+p=(x-3)(x^3-(a+b+c)x^2+(ab+bc+ca)x-abc)$

$\therefore 3^4-3^3-7(2)^2+3+p=0$

$p=6$[/TD]
[TD]$x^4+5x^3+5x^2-5x+q=(x+3)(x^3-(a+b+c)x^2+(ab+bc+ca)x-abc)$

$\therefore (-3)^4+5(-3)^3+5(-3)^2-5(-3)+q=0$$q=-6$[/TD]
[/TR]
[/TABLE]

And therefore, $162p-172q=162(6)-172(-6)=2004$.
 
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