## Tuesday, March 21, 2006

### Para-Para-Parameter. Topic: Algebra. Level: Olympiad.

Problem: (360 Problems For Mathematical Contests - 1.1.59) Solve the equation

$x^4-(2m+1)x^3+(m-1)x^2+(2m^2+1)x+m = 0$

where $m$ is a real parameter.

Solution: Well, this equation just screams to be factored... but doing so is the challenge. Instead of a quartic in $x$ let's write it as a quadratic in $m$.

$(2x)m^2-(x-1)(2x^2+x+1)m+x(x-1)^2(x+1) = 0$

after everything is factored fully. Well this looks a little more manageable, so we guess for factors. Noticing the $x-1$ term on the $m$, we can see that $x-1$ must appear in both factors, so the square must be split. After some work, we come up with

$[2xm-(x+1)(x-1)][m-x(x-1)] = 0$

$(x^2-2mx-1)(x^2-x-m) = 0$.

Solving this, we have either

$x^2-2mx-1 = 0 \Rightarrow x = m \pm \sqrt{m^2+1}$

or

$x^2-x-m = 0 \Rightarrow x = \frac{1 \pm \sqrt{1+4m}}{2}$,

both just from using the quadratic formula. Hence these are our solutions for $x$ in terms of the parameter $m$. QED.

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Comment: This technique of switching the variable is very useful because solving quadratics is much simpler than higher degree polynomials. Problems will often give the equation in a form like this one with one variable of a large degree and another of degree two. It's possible to discover the factorization by looking at the fourth degree polynomial, but much more difficult.

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Practice Problem: (360 Problems For Mathematical Contests - 1.1.60) Solve the equation

$x^{2n}+a_1x^{2n-1}+\cdots+a_{2n-2}x^2-2nx+1 = 0$

if all of its roots are positive reals. [Note: Original problem omitted positive, but I think it's necessary from looking at the solution.]

#### 1 comment:

1. Practice Problem: ARML 2005 Individual #8 :D