Solve the equation $x^{4} - 8 x^{3} + 24 x^{2} - 32 x + 20 = 0$ if $3 + i$ is a root of it.

Open in App

Solution

Since $3 + i$ is a root, $3 - i$ is another root.
Sum of the roots $= 3 + i + 3 - i = 6$
Product of the roots $= (3 + i) (3 - i) = 3^{2} + 1^{2} = 10$
The equation with $3 + i$ and $3 - i$ as roots is
$x^{2} - (6) x + 10 = 0$
(i.e.), $x^{2} - 6 x + 10 = 0$
$\Rightarrow x^{4} - 8 x^{3} + 24 x^{2} - 32 x + 20 = (x^{2} - 6 x + 10) (x^{2} + p x + 2)$
Equating co-efficient of $x$
$\Rightarrow - 12 + 10 p = - 32$
$\Rightarrow 10 p = - 20$
$\Rightarrow p = - 2$
$∴$ the other factor is $x^{2} - 2 x + 2$
Solving $x^{2} - 2 x + 2 = 0$ , we get
$a = 1$ , $b = - 2$ , $c = 2$
Therefore, $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
$= \frac{+ 2 \pm \sqrt{4 - 8}}{2}$
$= \frac{2 \pm 2 i}{2}$
$= \frac{2 (1 \pm i)}{2}$
$= 1 \pm i$

Suggest Corrections

Similar questions

x^{4} + 8 x^{3} + 9 x^{2} - 8 x - 10 = 0

x^{4} - 8 x^{3} + a x^{2} - b x + 16 = 0

x^{4} - 8 x^{3} + b x^{2} + c x + 16 = 0

x^{4} - 4 x^{2} + 8 x + 35 = 0

2 + \sqrt{3} i

x^{4} - 8 x^{3} + x^{2} - x + 3 = 0

Join BYJU'S Learning Program