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?

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- Nov 19th 2012, 12:32 PMalgebra123Real roots of unity
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? - Nov 19th 2012, 12:50 PMDevenoRe: Real roots of unity
your equation doesn't look quite right.....

- Nov 19th 2012, 12:53 PMalgebra123Re: Real roots of unity
Not quite sure what happened, but I've changed it now.

- Nov 19th 2012, 01:26 PMPlatoRe: Real roots of unity
- Nov 19th 2012, 02:12 PMDevenoRe: Real roots of unity
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:

each cubic can be factored again:

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:

so these have no real roots. - Nov 19th 2012, 05:56 PMProve ItRe: Real roots of unity
The easy way is to remember that in the real numbers has two solutions if is even and , and has one solution if is odd.