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

[anemone]

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 $162p-172q$.

 

[mente oscura]

Re: Evaluate 162p-172q

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.

Spoiler

I have come to:

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

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

Regards.

 

[Opalg]

Re: Evaluate 162p-172q

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]

 

[anemone]

Re: Evaluate 162p-172q

Hello.

Spoiler

I have come to:

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

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

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.:)

[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:

Spoiler

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