Q.27. Polynomial equation.

Question 27.
One solution of x^3 + (2 - i)x^2 + (- 4 - 3i)x + (1 + i) = 0 is 1 + i. Find all positive real solutions.

Answer 27.
x^3 +(2-i)x^2 +(-4-3i)x +(1+i)=0
=> (x^3 + 2x^2 - 4x +1) - i (x^2 + 3x - 1) = 0

For real x, both real and imaginary parts must be zero, i.e.,
x^3 + 2x^2 - 4x + 1 = 0
=> x^3 - x^2 + 3x^2 - 3x - x + 1 = 0
=> x^2(x - 1) + 3x(x - 1) - (x - 1) = 0
=> (x -1) (x^2 + 3x - 1) = 0 ... ( 1 ) and
x^2 + 3x - 1 = 0 ... ... ... ... ( 2 )
For both these equations to be satisfied simultaneously,
x^2 + 3x - 1 = 0
=> x = (1/2) [ -3 ± √(13) ]
Since only positive solution is required, answer is
x = (1/2) [ - 3 + √(13) ]

Link to YA!