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- Feb 14, 2012

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

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

Evaluate $162p-172q$.

- Thread starter anemone
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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,909

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

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

Evaluate $162p-172q$.

- Nov 29, 2013

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Hello.

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

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

Evaluate $162p-172q$.

[tex]p=-q \rightarrow{}162p-172q=334p=-334q[/tex]

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

Regards.

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- Feb 7, 2012

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

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

Evaluate $162p-172q$.

$3^4 - 3^3 - 7\cdot3 + 3 + p = 0$, 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).$3^4 - 5\cdot3^3 + 5\cdot3 + 5\cdot3 + q = 0$, from which $q=-6$.

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- Feb 14, 2012

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HeyHello.

[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,$3^4 - 3^3 - 7\cdot3 + 3 + p = 0$, 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 $q=-6$.

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:

$x^4-x^3-7x^2+x+p=(x-3)(x^3-(a+b+c)x^2+(ab+bc+ca)x-abc)$ $\therefore 3^4-3^3-7(2)^2+3+p=0$ $p=6$ | $x^4+5x^3+5x^2-5x+q=(x+3)(x^3-(a+b+c)x^2+(ab+bc+ca)x-abc)$ $\therefore (-3)^4+5(-3)^3+5(-3)^2-5(-3)+q=0$$q=-6$ |

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