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About LinkBacksi am having problems understanding the roots
have the equation:
given that x = 4 is a real solution for f(x) = x^3 - 4x^2 + x - 4, find one of the complex roots?
and the steps to find a cube root of -8?
Since x = 4 is a solution, that means (x - 4) is a factor. Long divide to get the quadratic factor, and factorise that over the complex numbers to evaluate the other two solutions (they will be complex conjugates).Originally Posted by tman
To find...
Solve each of those two equations to find the three cube roots of -8.
an only slighter easier way:
suppose we already know how to find the 3 cube roots of 1 (by, say, using trigonometry), call these: 1, ω, and ω2 (=).
(it should be easy to see that if ω3 = 1, and ω ≠ 1, then (ω2)3 = (ω3)2 = (1)(1) = 1, as well).
(we can find these by factoring x3 - 1 = (x - 1)(x2 + x + 1), also).
then (x + 2)(x + 2ω)(x + 2ω2) = (x + 2)(x2 + (2ω + 2ω2)x + 4ω3)
= (x + 2)(x2 + 2(ω + ω2)x + 4), and we know from above that: ω + ω2 = -1, so:
= (x + 2)(x2 - 2x2 + 4) = x3 + 8
so the three cube roots of -8 are -2,-2ω, and -2ω2.
I wouldn't say that is easier, it seems like more work than the method I gave. Each to their ownan only slighter easier way:
suppose we already know how to find the 3 cube roots of 1 (by, say, using trigonometry), call these: 1, ω, and ω2 (=
(it should be easy to see that if ω3 = 1, and ω ≠ 1, then (ω2)3 = (ω3)2 = (1)(1) = 1, as well).
(we can find these by factoring x3 - 1 = (x - 1)(x2 + x + 1), also).
then (x + 2)(x + 2ω)(x + 2ω2) = (x + 2)(x2 + (2ω + 2ω2)x + 4ω3)
= (x + 2)(x2 + 2(ω + ω2)x + 4), and we know from above that: ω + ω2 = -1, so:
= (x + 2)(x2 - 2x2 + 4) = x3 + 8
so the three cube roots of -8 are -2,-2ω, and -2ω2.
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