Difference between revisions of "2003 AIME II Problems/Problem 9"

(Solution)
(Solution)
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Now <cmath>P(z_2)+P(z_1)+P(z_3)+P(z_4)=z_1^2-z_1+1+z_2^2-z_2+1+z_3^2-z_3+1+z_4^2-z_4+1</cmath>
 
Now <cmath>P(z_2)+P(z_1)+P(z_3)+P(z_4)=z_1^2-z_1+1+z_2^2-z_2+1+z_3^2-z_3+1+z_4^2-z_4+1</cmath>
  
Now by Vieta's we know that <math>-z_4-z_3-z_2-z_1=-1</math>
+
Now by Vieta's we know that <math>-z_4-z_3-z_2-z_1=-1</math>,
So by Newton Sums we can find <math>z_1^2+z_2^2+z_3^2+z_4^2</math>
+
so by Newton Sums we can find <math>z_1^2+z_2^2+z_3^2+z_4^2</math>
  
 
<math>a_ns_2+a_{n-1}s_1+2a_{n-2}=0</math>
 
<math>a_ns_2+a_{n-1}s_1+2a_{n-2}=0</math>

Revision as of 00:36, 5 July 2015

Problem

Consider the polynomials $P(x) = x^{6} - x^{5} - x^{3} - x^{2} - x$ and $Q(x) = x^{4} - x^{3} - x^{2} - 1.$ Given that $z_{1},z_{2},z_{3},$ and $z_{4}$ are the roots of $Q(x) = 0,$ find $P(z_{1}) + P(z_{2}) + P(z_{3}) + P(z_{4}).$

Solution

${{Q(z_1)=0}}$ therefore $z_1^4-z_1^3-z_1^2-1=0$

therefore \[-z_1^3-z_1^2=-z_1^4+1.\]

Also \[z_1^4-z_1^3-z_1^2=1\]

So \[z_1^6-z_1^5-z_1^4=z_1^2\]

So in \[P(z_1)=z_1^6-z_1^5-z_1^3-z_1^2-z_1\]

Since $-z_1^3-z^2=-z_1^4+1$ and $z_1^6-z_1^5-z_1^4=z_1^2$

\[P(z_1)=z_1^6-z_1^5-z_1^3-z_1^2-z_1\] can now be \[P(z_1)=z_1^2-z_1+1\]

Now this also follows for all roots of $Q(x)$ Now \[P(z_2)+P(z_1)+P(z_3)+P(z_4)=z_1^2-z_1+1+z_2^2-z_2+1+z_3^2-z_3+1+z_4^2-z_4+1\]

Now by Vieta's we know that $-z_4-z_3-z_2-z_1=-1$, so by Newton Sums we can find $z_1^2+z_2^2+z_3^2+z_4^2$

$a_ns_2+a_{n-1}s_1+2a_{n-2}=0$

$(1)(s_2)+(-1)(1)+2(-1)=0$

$s_2-1-2=0$

$s_2=3$

So finally $P(z_2)+P(z_1)+P(z_3)+P(z_4)=3+4-1=\boxed{6}$

See also

2003 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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