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- Feb 14, 2012
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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 $162p-172q$.
$x^4-x^3-7x^2+x+p=0$
$x^4+5x^3+5x^2-5x+q=0$
Evaluate $162p-172q$.
Hello.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 $162p-172q$.
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 $162p-172q$.
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.Hello.
I have come to:
[tex]p=-q \rightarrow{}162p-172q=334p=-334q[/tex]
If it is that, the solution sought, I show the test.
Regards.
Awesome! Your approach is a great one and thanks for participating again, Opalg!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 $p=6$,Therefore $162p - 172q = 6(162+172) = 2004$ (presumably this problem was set in that year).
$3^4 - 5\cdot3^3 + 5\cdot3 + 5\cdot3 + q = 0$, from which $q=-6$.
$x^4-x^3-7x^2+x+p=(x-3)(x^3-(a+b+c)x^2+(ab+bc+ca)x-abc)$ $\therefore 3^4-3^3-7(2)^2+3+p=0$ $p=6$ | $x^4+5x^3+5x^2-5x+q=(x+3)(x^3-(a+b+c)x^2+(ab+bc+ca)x-abc)$ $\therefore (-3)^4+5(-3)^3+5(-3)^2-5(-3)+q=0$$q=-6$ |