How many real roots does the equation (z-2)^12 = 6 have?
i have no idea how to go about this, can someone help me?
note that z - 2 real implies that z = (z - 2) + 2 is real, and vice versa.
so let u = z - 2, so we want the number of real roots to u^{12} = 6.
that is, we're looking for real factors of u^{12} - 6.
now u^{12} - 6 = (u^{6} + √6)(u^{6} - √6).
notice that u^{6} + √6 > 0 for any real u, so we don't get any solutions there. so any REAL roots of u^{12} - 6 must be real roots of u^{6} - √6.
again, we have:
$\displaystyle u^6 - \sqrt{6} = (u^3 + \sqrt[4]{6})(u^3 - \sqrt[4]{6})$
each cubic can be factored again:
$\displaystyle u^3 + \sqrt[4]{6} = (u + \sqrt[12]{6})(u^2 - \sqrt[12]{6}u + \sqrt[6]{6})$
$\displaystyle u^3 - \sqrt[4]{6} = (u - \sqrt[12]{6})(u^2 + \sqrt[12]{6}u + \sqrt[6]{6})$
this gives us two real roots u (to get z we'll have to add 2 to u). what about the quadratic factors?
let's look at their discriminants:
b^{2} - 4c (the same for both quadratics, since they only differ in the sign for b).
we have:
$\displaystyle (\sqrt[12]{6})^2 - 4(\sqrt[6]{6}) = \sqrt[6]{6} - 4\sqrt[6]{6} = (-3)\sqrt[6]{6} < 0$ so these have no real roots.
The easy way is to remember that in the real numbers $\displaystyle \displaystyle \begin{align*} x^a = b \end{align*}$ has two solutions if $\displaystyle \displaystyle \begin{align*} a \end{align*}$ is even and $\displaystyle \displaystyle \begin{align*} b \geq 0 \end{align*}$, and has one solution if $\displaystyle \displaystyle \begin{align*} a \end{align*}$ is odd.
$\displaystyle \displaystyle \begin{align*} (z - 2)^{12} &= 6 \\ z - 2 &= \pm \sqrt[12]{6} \\ z &= 2 \pm \sqrt[12]{6} \end{align*}$