How do you solve x^3=4x?

Jun 9, 2016

$x = 0 , \text{ or "x= -2," or } x = 2$

Explanation:

the tempting method is just to divide by $x$ which gives
${x}^{2} = 4$

However in dividing by $x$ we have made an assumption that $x$ is not 0, in which case it would not be allowed.

Make the equation = 0: $\text{ } {x}^{3} - 4 x = 0$

Factorise fully: $\text{ } x \left({x}^{2} - 4\right) = 0$

So, $x \left(x + 2\right) \left(x - 2\right) = 0$

Each factor can be equal to 0, so
$x = 0 , \text{ or "x= -2," or } x = 2$

An equation with ${x}^{3}$ should have 3 solutions, which we have.