# 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

Sep 17, 2017

$- \frac{3}{2} + \frac{\sqrt{13}}{2}$

#### Explanation:

${x}^{3} + \left(2 - i\right) {x}^{2} + \left(- 4 - 3 i\right) x + \left(1 + i\right)$

$= {x}^{3} - \left(1 + i\right) {x}^{2} + 3 {x}^{2} + \left(- 4 - 3 i\right) x + \left(1 + i\right)$

$= {x}^{3} - \left(1 + i\right) {x}^{2} + 3 {x}^{2} - 3 \left(1 + i\right) x - x + \left(1 + i\right)$

$= {x}^{2} \left(x - \left(1 + i\right)\right) + 3 x \left(x - \left(1 + i\right)\right) - 1 \left(x - \left(1 + i\right)\right)$

$= \left({x}^{2} + 3 x - 1\right) \left(x - \left(1 + i\right)\right)$

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

$x = \frac{- 3 \pm \sqrt{{3}^{2} - 4 \left(1\right) \left(- 1\right)}}{2 \left(1\right)}$

$\textcolor{w h i t e}{x} = - \frac{3}{2} \pm \frac{\sqrt{13}}{2}$

So the positive real root is:

$- \frac{3}{2} + \frac{\sqrt{13}}{2}$