Anonymous
Anonymous asked in Science & MathematicsMathematics · 4 weeks ago

# math exercise?

Solve: x^3 - 6x + 8 = 0

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• Ian H
Lv 7
4 weeks ago

For x^3 – 3px – 2q = 0 with q^2 > p^3 there is one real root

x^3 - 6x + 8 = 0 has p = 2, q = -4, so q^2 > p^3

Substitution x = s + p/s and multiplying by s^3 yields the quadratic of cubes

(s^3)^2 – 2q(s^3) + p^3 = 0

s^3 = q ±√(q^2 – p^3) = -4 ±√(8) ~ -1.1715728752538

Selecting the real cube root, s ~ -1.0542002205047

x = s + 2/s ~ -2.9513730355915..., which we may call r

That is the real root; but every cubic has 3 roots.

The remaining quadratic must be of the form x^2 + rx - 8/r

With solutions x = [-r ±√(r^2 +32/r)]/2

In this example √[-(r^2 + 32/r)]/2 = + 0.730035681602

So the complex conjugate roots are

x = 1.4756865177958 ± 0.730035681602 i

• 4 weeks ago

Could you perhaps have meant for that to be a quadratic rather than a cubic?

In the off-chance, you have a typo, I'll show you how to factor the quadratic:

x² - 6x + 8 = 0

(x - 2)(x - 4) = 0

Using the zero product rule, if ab=0, then a=0 or b=0:

x - 2 = 0

x = 2

or

x - 4 = 0

x = 4