Question

Find the modulus of the sum of all solutions to $\left(x^2+3 x+1\right)^{x^2+2 x-8}=1$.

Answer 1

$\left(x^2+3 x+1\right)^{x^2+2 x-8}=1$ is true if
(i) the exponent $=0$ and base is not zero,
(ii) the base $=1$, or
(iii) the base $=-1$ and the exponent is even.
(i) $x ^2+2 x -8=0 \Rightarrow( x +4)( x -2)=0 \Rightarrow x =-4,2$
(ii) $x^2+3 x+1=1 \Rightarrow x^2+3 x=0 \Rightarrow x(x+3)=0 \Rightarrow x=-3,0$
(iii) $x^2+3 x+1=-1 \Rightarrow x^2+3 x+2=0 \Rightarrow(x+1)(x+2)=0 \Rightarrow x=-2,-1$.
Now, check the exponent. If $x=-2 \Rightarrow x^2+2 x-8$ is even; if $x=-1 \Rightarrow x^2+2 x-8$ is odd; therefore case (iii) produces the solution -2 only.
The sum of all the solutions $=-4+2+(-3)+0+(-2)=-7$
Hence modulus is 7 .