Solve the equation : $x (x - 1) (x + 2) (x - 3) + 8 = 0.$

Open in App

Solution

$x (x - 1) (x + 2) (x - 3) + 8 = 0$
$\Rightarrow (x^{2} - x) (x^{2} + (2 - 3) x + 2 \times - 3) + 8 = 0$
$\Rightarrow (x^{2} - x) (x^{2} - x - 6) + 8 = 0$
$\Rightarrow x^{4} - x^{3} - 6 x^{2} - x^{3} + x^{2} + 6 x + 8 = 0$
$\Rightarrow x^{4} - 2 x^{3} - 5 x^{2} + 6 x + 8 = 0$
Let $f (x) = x^{4} - 2 x^{3} - 5 x^{2} + 6 x + 8 = 0$
Let $x = - 1 \Rightarrow f (- 1) = {(- 1)}^{4} - 2 {(- 1)}^{3} - 5 {(- 1)}^{2} + 6 (- 1) + 8 = 0$
$= 1 + 2 - 5 - 6 + 8 = 0$
$∴ x = - 1$ is one root of the above equation.

Suggest Corrections

Similar questions

x :

2 (x + 1) (x + 3) + 8 = (2 x + 1) (x + 5)

∣ ∣ ∣ ∣ \begin{matrix} x + 2 x + 6 x - 1 \end{matrix} \begin{matrix} x + 6 x - 1 x + 2 \end{matrix} \begin{matrix} x - 1 x + 2 x + 6 \end{matrix} ∣ ∣ ∣ ∣ = 0

\sqrt{\frac{x}{x - 1}} + \sqrt{\frac{x - 1}{x}} = \frac{13}{6}, x \neq 0, 1

x(x + 1)(x + 2)(x + 3) ÷ x(x + 1) = ______

$\frac{x (x + 1) (x + 2) (x + 3)}{x (x + 1)}$ = ______

Join BYJU'S Learning Program