Question

One solution of `x^3+(2-i)x^2+(-4-3i)x+(1+i)=0` is `x=1+i`. Find the only positive real solution for `x`?

Hint: `1-i` is not necessarily a root because the coefficients are imaginary

Answer 1

`-3/2+sqrt(13)/2`

Explanation:

`x^3+(2-i)x^2+(-4-3i)x+(1+i)`

`=x^3-(1+i)x^2+3x^2+(-4-3i)x+(1+i)`

`=x^3-(1+i)x^2+3x^2-3(1+i)x-x+(1+i)`

`=x^2(x-(1+i))+3x(x-(1+i))-1(x-(1+i))`

`=(x^2+3x-1)(x-(1+i))`

So the other two roots are the roots of `x^2+3x-1 = 0`, which we can find using the quadratic formula:

`x = (-3+-sqrt(3^2-4(1)(-1)))/(2(1))`

`color(white)(x) = -3/2+-sqrt(13)/2`

So the positive real root is:

`-3/2+sqrt(13)/2`