Suppose
$$(x^2 - 5)^{8} (x+1)^{-16} = 1$$
How do we solve for all values of $x$?
I think we can equate the bases to $1$ in order to get $$1 \times 1 = 1$$
$$x^2-5 = 1, x^2-5=-1$$
or
$$x + 1 = 1, x + 1 = -1$$
Could you assist me with this?
$\endgroup$3 Answers
$\begingroup$$(x^2 - 5)^{8} (x+1)^{-16} = 1$ means
$(x^2 - 5)^{8} = (x+1)^{16}$.
Assuming you only want real and not complex answers then
$ \sqrt[8]{(x^2 - 5)^{8}} = \sqrt[8]{(x+1)^{16}}$
$|x^2 - 5| = (x+1)^2$
or either A) $x^2 - 5 = (x+1)^2$ or B) $5- x^2 = (x+1)^2$
A) $x^2 - 5 = x^2 + 2x + 1$
$2x = -6$
$x = -3$.
Verification: $(x^2 - 5)^8(x+1)^{-16}=\frac{(9-5)^8}{(-2)^{16}}= \frac {4^8}{2^{16}}=\frac {2^{16}}{2^{16}} = 1$.
B)$5- x^2 = (x+1)^2$
$5-x^2 = x^2 + 2x + 1$
$2x^2 + 2x -4 = 0$
$x^2 + x -2 = 0$
$(x +2)(x-1) = 0$ so either C) $x+2 = 0$ and $x=-2$ or $x-1=0$ and $x = 1$.
So solutions are $x = -3; x = -2;$ or $x = 1$.
Verification:
$(x^2 - 5)^8(x+1)^{-16} = \frac {(4-5)^8}{(-2 + 1)^{16}}= \frac {(-1)^8}{(-1)^{16}} = \frac 11 = 1$. so $x = -2$ is a solution.
$(x^2 - 5)^8(x+1)^{-16} = \frac {(1-5)^8}{(1 + 1)^{16}}= \frac {(-4)^8}{(2)^{16}} = \frac {4^8}{4^8} = 1$. so $x = 1$ is a solution.
$\endgroup$ $\begingroup$We have $(x^2-5)^8(x+1)^{-16}=1$, so $\left(\frac{x^2-5}{(x+1)^2}\right)^8=1$, so $\frac{x^2-5}{(x+1)^2}$ is an $8$'th root of unity. Assuming you need to solve over the reals, we get $\frac{x^2-5}{(x+1)^2}\in\{-1,1\}$. Fixing for $-1$ or $1$, you can multiply both sides by $(x+1)^2$ to get a quadratic equation.
$\endgroup$ $\begingroup$$(x^2-5)^8=(x+1)^{16}$
So
$[(x^2-5)^4-(x+1)^8] [(x^2-5)^4+(x+1)^8]=0$
If and only if
$[(x^2-5)^4-(x+1)^8]=0$
If and only if
$[(x^2-5)^2-(x+1)^4]=0$
If and only if
$[(x^2-5)-(x+1)^2] [(x^2-5)+(x+1)^2]=0$
So
$[-2x-6][2x^2+2x-4]=0$
$\endgroup$
