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Difference between revisions of "2010 AMC 12A Problems/Problem 21"

(Created page with '== Problem 21 == The graph of <math>y=x^6-10x^5+29x^4-4x^3+ax^2</math> lies above the line <math>y=bx+c</math> except at three values of <math>x</math>, where the graph and the l…')
 
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== Problem 21 ==
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== Problem ==
 
The graph of <math>y=x^6-10x^5+29x^4-4x^3+ax^2</math> lies above the line <math>y=bx+c</math> except at three values of <math>x</math>, where the graph and the line intersect. What is the largest of these values?
 
The graph of <math>y=x^6-10x^5+29x^4-4x^3+ax^2</math> lies above the line <math>y=bx+c</math> except at three values of <math>x</math>, where the graph and the line intersect. What is the largest of these values?
  
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Suppose we let <math>p</math>, <math>q</math>, and <math>r</math> be the roots of this function, and let <math>x^3-ux^2+vx-w</math> be the cubic polynomial with roots <math>p</math>, <math>q</math>, and <math>r</math>.
 
Suppose we let <math>p</math>, <math>q</math>, and <math>r</math> be the roots of this function, and let <math>x^3-ux^2+vx-w</math> be the cubic polynomial with roots <math>p</math>, <math>q</math>, and <math>r</math>.
  
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<cmath>\begin{align*}(x-p)(x-q)(x-r) &= x^3-ux^2+vx-w\\
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<math>(x-p)(x-q)(x-r) = x^3-ux^2+vx-w</math>
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(x-p)^2(x-q)^2(x-r)^2 &= x^6-10x^5+29x^4-4x^3+ax^2-bx-c = 0\\
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\sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c} &= x^3-ux^2+vx-w = 0\end{align*}</cmath>
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<math>(x-p)^2(x-q)^2(x-r)^2 = x^6-10x^5+29x^4-4x^3+ax^2-bx-c = 0</math>
 
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<math>\sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c} = x^3-ux^2+vx-w = 0</math>
 
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In order to find <math>\sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c}</math> we must first expand out the terms of <math>(x^3-ux^2+vx-w)^2</math>.
 
In order to find <math>\sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c}</math> we must first expand out the terms of <math>(x^3-ux^2+vx-w)^2</math>.
  
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<cmath>(x^3-ux^2+vx-w)^2</cmath>
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<math>(x^3-ux^2+vx-w)^2</math>
 
 
<math>= x^6-2ux^5+(u^2+2v)x^4-(2uv+2w)x^3+(2uw+v^2)x^2-2vwx+w^2</math>
 
<math>= x^6-2ux^5+(u^2+2v)x^4-(2uv+2w)x^3+(2uw+v^2)x^2-2vwx+w^2</math>
  
 
[Quick note: Since we don't know <math>a</math>, <math>b</math>, and <math>c</math>, we really don't even need the last 3 terms of the expansion.]
 
[Quick note: Since we don't know <math>a</math>, <math>b</math>, and <math>c</math>, we really don't even need the last 3 terms of the expansion.]
  
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<math>2u = 10</math>
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<cmath>\begin{align*}&2u = 10\\
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u^2+2v &= 29\\
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2uv+2w &= 4\\
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u &= 5\\
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v &= 2\\
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w &= -8\\
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&\sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c} = x^3-5x^2+2x+8\end{align*}</cmath>
  
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<math>u^2+2v = 29</math>
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All that's left is to find the largest root of <math>x^3-5x^2+2x+8</math>.
  
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<math>2uv+2w = 4</math>
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<cmath>\begin{align*}&x^3-5x^2+2x+8 = (x-4)(x-2)(x+1)\\
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&\boxed{\textbf{(A)}\ 4}\end{align*}</cmath>
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<math>u = 5</math>
 
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<math>v = 2</math>
 
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<math>w = -8</math>
 
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<math>\sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c} = x^3-5x^2+2x+8</math>
 
−
 
 
−
 
 
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All that's left is to find the largest root of <math>x^3-5x^2+2x+8</math>.
 
  
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<math>x^3-5x^2+2x+8 = (x-4)(x-2)(x+1)</math>
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== See also ==
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{{AMC12 box|year=2010|num-b=20|num-a=22|ab=A}}
  
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<math>\boxed{\textbf{(A)}\ 4}</math>
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[[Category:Intermediate Algebra Problems]]

Revision as of 22:34, 25 February 2010

Problem

The graph of $y=x^6-10x^5+29x^4-4x^3+ax^2$ lies above the line $y=bx+c$ except at three values of $x$, where the graph and the line intersect. What is the largest of these values?

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$

Solution

Solution

The $x$ values in which $y=x^6-10x^5+29x^4-4x^3+ax^2$ intersect at $y=bx+c$ are the same as the zeros of $y=x^6-10x^5+29x^4-4x^3+ax^2-bx-c$.

Since there are $3$ zeros and the function is never negative, all $3$ zeros must be double roots because the function's degree is $6$.

Suppose we let $p$, $q$, and $r$ be the roots of this function, and let $x^3-ux^2+vx-w$ be the cubic polynomial with roots $p$, $q$, and $r$.

\begin{align*}(x-p)(x-q)(x-r) &= x^3-ux^2+vx-w\\ (x-p)^2(x-q)^2(x-r)^2 &= x^6-10x^5+29x^4-4x^3+ax^2-bx-c = 0\\ \sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c} &= x^3-ux^2+vx-w = 0\end{align*}

In order to find $\sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c}$ we must first expand out the terms of $(x^3-ux^2+vx-w)^2$.

\[(x^3-ux^2+vx-w)^2\] $= x^6-2ux^5+(u^2+2v)x^4-(2uv+2w)x^3+(2uw+v^2)x^2-2vwx+w^2$

[Quick note: Since we don't know $a$, $b$, and $c$, we really don't even need the last 3 terms of the expansion.]

\begin{align*}&2u = 10\\ u^2+2v &= 29\\ 2uv+2w &= 4\\ u &= 5\\ v &= 2\\ w &= -8\\ &\sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c} = x^3-5x^2+2x+8\end{align*}

All that's left is to find the largest root of $x^3-5x^2+2x+8$.

\begin{align*}&x^3-5x^2+2x+8 = (x-4)(x-2)(x+1)\\ &\boxed{\textbf{(A)}\ 4}\end{align*}

See also

2010 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
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