Find all possible real values of $a$ for which the system of equations \[\{\begin{array}{cc}x +y +z=0\\\text{ } \\ xy+yz+azx=0\end{array}\] has exactly one solution.
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Tags: algebra, system of equations, algebra unsolved
modularmarc101
23.01.2011 05:33
Notice that no matter what $a$ is, $(x,y,z) = (0,0,0)$ will always be a solution, so we want this to be our only solution. First, substitute $y = -(x+z)$ into the second equation and it becomes
$-(x+z)^2 + azx = 0$
or
$x^2 - xz(a - 2) + z^2 = 0$.
Let $t = \frac{x}{z}$, where $z \neq 0$, then the equation above is equivalent to
$t^2 - t(a-2) + 1 = 0$.
Since we want $(0,0,0)$ to be our only solution, then this equation must have no solution, so
$\Delta = (a-2)^2 - 4 < 0$
$\implies a(a-4) < 0$
$\implies 0 < a < 4$
So we've reduced the possible values for $a$ down to the interval $(0,4)$.
SCP
30.10.2011 18:46
modularmarc101 wrote: So we've reduced the possible values for $a$ down to the interval $(0,4)$. [/hide] All of these values have to work, so it is the whole solution, isn't it?