$x^{3} (x^{2} - 5) = - 4 x$
If $x > 0$ , what is one possible solution to the equation above?

1

2

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1

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2

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Solution

The correct option is A $1$ or $2$
$x^{3} (x^{2} - 5) = - 4 x$
$\Rightarrow x^{5} - 5 x^{3} + 4 x = 0$
$\Rightarrow x (x^{4} - 5 x^{2} + 4) = 0$
$\Rightarrow x = 0$ or $x^{4} - 5 x^{2} + 4 = 0$
But given $x > 0$
$∴ x^{4} - 5 x^{2} + 4 = 0$
$x^{4} - 4 x^{2} - x^{2} + 4 = 0$
$\Rightarrow (x^{2} - 4) (x^{2} - 1) = 0$
$x^{2} - 4 = 0$ or $x^{2} - 1 = 0$
$x^{2} - 4 = 0 ⟹ x = \pm 2$
$x^{2} - 1 = 0 ⟹ x = \pm 1$
Since it is given $x > 0$
$x = 1$ or $x = 2$

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