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

1 Answer
Sep 17, 2017

-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