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
Hint:
1 Answer
Sep 17, 2017
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 = (-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