Difference between revisions of "2007 Alabama ARML TST Problems/Problem 14"

Latest revision as of 07:09, 26 January 2009

Problem

Find the sum of the real roots of

$x^6+145x^4-2007x^3+145x^2+1=0.$

Solution

Step 1: using the symmetry

Let $p(x)=x^6+145x^4-2007x^3+145x^2+1$

The polynomial $p$ is symmetric. Each symmetric polynomial has the following property: for $x\not=0$ , $x$ is a root if and only if $\frac 1x$ is.

To see why this is true, assume that we have a $x\not=0$ such that $p(x)=x^6+145x^4-2007x^3+145x^2+1=0$ . We can now divide both sides by $x^6$ . We get:

$1 + \frac{145}{x^2} - \frac{2007}{x^3} + \frac{145}{x^4} + \frac{1}{x^6} = 0$

But the left hand side is just $p\left(\frac 1x\right)$ . Thus also $p\left(\frac 1x\right)=0$ .

Step 2: multiplying three quadratic terms

Our polynomial has six complex roots: $\alpha$ , $1/\alpha$ , $\beta$ , $1/\beta$ , $\gamma$ , and $1/\gamma$ . It can be expressed as:

$p(x) = (x-\alpha)(x-1/\alpha)(x-\beta)(x-1/\beta)(x-\gamma)(x-1/\gamma)$

Let $a=\alpha+1/\alpha$ , $b=\beta+1/\beta$ , and $c=\gamma+1/\gamma$ . We can then write:

$p(x) = (x^2 - ax + 1)(x^2 - bx + 1)(x^2 - cx + 1)$

We can now compare the coefficients of $x$ , $x^2$ and $x^3$ in this expression. (Comparing other coefficients is not necessary due to symmetry.) We get the following equations:

$-a-b-c=0$
$3+ab+ac+bc=145$
$-abc-2a-2b-2c=-2007$

Simplifying, we get:

$a+b+c=0$
$ab+ac+bc=142$
$abc=2007$

Using Vieta's formulas, this means that $a$ , $b$ and $c$ are the roots of the polynomial $q(x)=x^3 - 0x^2 + 142x - 2007$ .

Step 3: Finding a root of q

Obviously, $q$ is increasing for $x>0$ , and $q(0)<0$ . Thus $q$ has exactly one positive root. We can easily find that $q(9)=0$ . Thus WLOG $a=9$ .

(Note: Remembering that $a+b+c=0$ and $abc=2007$ , we now have $b+c=-a=-9$ and $bc=2007/9=223$ . From the first equation, let $c=-9-b$ . Substituting into the second one, we get $b(b+9)=-223$ . This obviously has no real solution, thus $b$ and $c$ are not real. But we don't need this observation.)

Step 4: Partially dividing p

We now know that $(x^2-9x+1)$ is a factor of $p$ . We can partially divide $p$ to get:

$p(x) = (x^2 - 9x + 1)(x^4 + 9x^3 + 225x^2 + 9x + 1)$

Step 5: Handling the remaining part of p

The polynomial $r(x)=x^4 + 9x^3 + 225x^2 + 9x + 1$ has no real roots.

To prove this, one can for example write:

$r(x) = \left(x^4 + 9x^3 + \frac{81}{4}\cdot x^2\right) + \left(\frac{81}{4}\cdot x^2 + 9x + 1\right) + \left(225-2\cdot\frac{81}{4}\right) x^2$

$r(x) = x^2\left(x+\frac92\right)^2 + \left(\frac{9x}2+1\right)^2 + \frac{369}2\cdot x^2$

Step 6: Summary

We found that $p(x)=(x^2 - 9x + 1)r(x)$ , where $r$ has no real roots. Thus all real roots of $p$ are the two real roots of $x^2 - 9x + 1$ , and their sum is obviously $\boxed{9}$ .

Note

  roots of p computed by Octave:
  -4.48006 + 14.30270i
  -4.48006 - 14.30270i
   8.88748 +  0.00000i
   0.11252 +  0.00000i
  -0.01994 +  0.06367i
  -0.01994 -  0.06367i

@@ Line 60: / Line 60: @@
 To prove this, one can for example write:
-<cmath>r(x) = (x^4 + 9x^3 + 81x^2) + (81x^2 + 9x + 1) + 63x^2 = x^2(x+9)^2 + (9x+1)^2 + 63x^2</cmath>
+<cmath>r(x) = \left(x^4 + 9x^3 + \frac{81}{4}\cdot x^2\right) + \left(\frac{81}{4}\cdot x^2 + 9x + 1\right) + \left(225-2\cdot\frac{81}{4}\right) x^2 </cmath>
+<cmath>r(x) = x^2\left(x+\frac92\right)^2 + \left(\frac{9x}2+1\right)^2 + \frac{369}2\cdot x^2</cmath>
 === Step 6: Summary ===

2007 Alabama ARML TST (Problems)
Preceded by: Problem 13	Followed by: Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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