If $α$ $alpha$ is a root of $x (2 - x) = 3$ $x(2-x) = 3$ then can you define a cubic equation with integer coefficients and roots including $- 2$ $-2$ and $α$ $alpha$ ?

Question

1199 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-18T07:48:39

Yes

Given:

$3 = x (2 - x) = 2 x - x^{2}$ $3 = x(2-x) = 2x-x^2$

Add $x^{2} - 2 x$ $x^2-2x$ to both ends to get

$x^{2} - 2 x + 3 = 0$ $x^2-2x+3 = 0$

Multiply both sides by $(x + 2)$ $(x+2)$ to get:

$0 = (x^{2} - 2 x + 3) (x + 2) = x^{3} - x + 6$ $0 = (x^2-2x+3)(x+2) = x^3-x+6$

Add $x - 6$ $x-6$ to both ends and transpose to get:

$x^{3} = x - 6$ $x^3 = x-6$

So the roots of this cubic equation are $x = - 2$ $x = -2$ and the roots of $x (2 - x) = 3$ $x(2-x) = 3$ , including $α$ $alpha$ .

If alphaαalpha is a root of x(2-x) = 3x(2−x)=3x(2-x) = 3 then can you define a cubic equation with integer coefficients and roots including -2−2-2 and alphaαalpha ?