Solve the equations:
$x^{4} + 4 x^{3} + 5 x^{2} + 2 x - 2 = 0$ , one root being $- 1 + \sqrt{- 1}$ .

Open in App

Solution

$p (x) = x^{4} + 4 x^{3} + 5 x^{2} + 2 x - 2 = 0$

If one of the roots id $(- 1 + i)$ then $(- 1 - i)$ is alos a root of $p (x)$

$\Rightarrow {x - (- 1 + i)} {x - (- 1 - i)}$ is a factor of $p (x)$

$\Rightarrow {(x + 1)}^{2} - {(i)}^{2}$ is a factor of $p (x)$

$\Rightarrow x^{2} + 2 x + 2$ is a factor of $p (x)$

Dividing $p (x)$ by $x^{2} + 2 x + 2$

$\Rightarrow p (x) = (x^{2} + 2 x - 1) (x^{2} + 2 x + 2) \Rightarrow (x^{2} + 2 x - 1) (x^{2} + 2 x + 2) \Rightarrow x^{2} + 2 x - 1 = 0 \Rightarrow x = \frac{- 2 \pm \sqrt{4 - 4 (1) (- 1)}}{2} \Rightarrow x = \frac{- 2 \pm 2 \sqrt{2}}{2} \Rightarrow x = - 1 \pm \sqrt{2}$

So the other two roots are $- 1 + \sqrt{2}$ and $- 1 - \sqrt{2}$

Suggest Corrections

Similar questions

x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 5 = 0

\sqrt{- 1}

Solve the equation $x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 5 = 0$ , given one root is $\sqrt{(- 1)}$ . Find the value of other three roots.

x^{5} - x^{4} + 8 x^{2} - 9 x - 15 = 0

\sqrt{3}

1 - 2 \sqrt{- 1}

The roots of the equation $x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 1 = 0$ are

4 x^{3} - 5 x^{2} - 6 x + 7 = 0

x = 1

Join BYJU'S Learning Program