Given that $(-2)$ is the solution of the equation :- $\frac{1}{3}mx = 5x + (-2)^2$ , find $(m^2 - 11m + 17)^{2007} .$

Question

Given that $(-2)$ is the solution of the equation :- $\frac{1}{3}mx = 5x + (-2)^2$ , find $(m^2 - 11m + 17)^{2007} .$

What I Tried :- We have :-

$5x + 4 - \frac{1}{3}mx = 0$

$\rightarrow 15x + 12 - mx = 0$

$\rightarrow 12x + 12 + 3x - mx = 0$

$\rightarrow 12(x + 1) + x(3 - m) = 0$

The Problem I am facing is, it's given that only $(-2)$ is $1$ of the solutions, but I can't say if $x = (-2)$ , or $m = (-2)$ .

If $x = (-2)$ ; $-12 + (-2)(3 - m) = 0$

$\rightarrow 2m - 18 = 0$ $\rightarrow m = 9.$

$m = (-2)$ can be a possibility as well.

Now, how to find $(m^2 - 11m + 17)^{2007}$ $?$ This seems like a huge expression and I am not trying to calculate it, of course no help from a calculator .

I am also not sure about the $(-2)$ part , and I guess my solution has got some issue . Please help me in this problem .

Thank You !

Answer 1

Since $x=-2$ is the solution of the equation then we have $15x+12-mx=15(-2)+12+2m=-18+2m=0$ and thus $m=9$ is the only solution for $x=-2$.

Then $(m^{2}-11m+17)^{2007}=(9^{2}-11(9)+17)^{2007}=(-1)^{2007}=-1.$