find a real solution for a complex root

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September 16th 2012, 08:05 AM
tman

find a real solution for a complex root

i 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?
September 16th 2012, 08:31 AM
Prove It

Re: find a real solution for a complex root

Quote:

Originally Posted by tman

i 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).

To find $\displaystyle \begin{align*} \sqrt[3]{-8} \end{align*}$ ...

$\displaystyle \begin{align*} z^3 &= -8 \\ z^3 + 8 &= 0 \\ (z + 2)\left( z^2 - 2z + 4 \right) &= 0 \\ z + 2 = 0 \textrm{ or } z^2 - 2z + 4 &= 0 \end{align*}$

Solve each of those two equations to find the three cube roots of -8.
September 16th 2012, 10:32 AM
Deveno

Re: find a real solution for a complex root

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 (= $\overline{\omega}$ ).

(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.
September 16th 2012, 12:01 PM
tman

Re: find a real solution for a complex root

you mean to factor out a x to get x^2-4x+4
then (x-2)(x-2)/x-4?
September 16th 2012, 08:50 PM
Prove It

Re: find a real solution for a complex root

Quote:

Originally Posted by tman

you mean to factor out a x to get x^2-4x+4
then (x-2)(x-2)/x-4?

I don't know who you're speaking to, but no, there is no factoring of a single x. You are going to long divide $\displaystyle \begin{align*} \frac{x^3 - 4x^2 + x - 4}{x - 4} \end{align*}$ .

The top is also able to be factorised by grouping...

$\displaystyle \begin{align*} x^3 - 4x^2 + x - 4 &= x^3 + x - 4x^2 - 4 \\ &= x\left( x^2 + 1 \right) - 4 \left( x^2 + 1 \right) \\ &= (x - 4) \left( x^2 + 1 \right) \\ &= (x - 4) (x - i)(x + i) \end{align*}$
September 16th 2012, 08:51 PM
Prove It

Re: find a real solution for a complex root

Quote:

Originally Posted by Deveno

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 (= $\overline{\omega}$ ).

(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 own :)