Question

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

Answer 1

Answer: 7

${\left(x^2+3x+1\right)}^{x^2+2x-8}=1$ 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) $x^2+2x-8=0\Rightarrow (x+4)(x-2)=0\Rightarrow x=-4,2$
(ii) $x^2+3x+1=1\Rightarrow x^2+3x=0\Rightarrow x(x+3)=0\Rightarrow x=-3,0$
(iii) $x^2+3x+1=-1\Rightarrow x^2+3x+2=0\Rightarrow (x+1)(x+2)=0\Rightarrow x=-2,-1$.
Now, check the exponent. If $x=-2\Rightarrow x^2+2x-8$ is even; if $x=-1\Rightarrow x^2+2x-8$ 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 .