Completing the square help

Question

The textbook gives this equation:

${12x^2 + 24x -8x = 0}$ with an answer of ${x = 0}$ or ${x = -{4\over3}}$

But I suspect it should be ${12x^2 + 24x -8 = 0}$

So in order to solve this, I would first isolate the x terms on one side of the equation by adding 8 to both sides:

${12x^2 + 24x = 8}$

I would then divide both sides by the coefficient of the ${x^2}$ or 12 in this case which gives:

${x^2 + 2x = {8\over12}}$

I then divide the coefficient of x by 2 and square the result and add it to both sides

${x^2 + 2x + 1 = {8\over12}}$

=> ${(x + 1)^2 = {8\over12}}$

=> ${x + 1 = \pm \sqrt{8\over12}}$

=> ${x + 1 = \pm \sqrt{2\over3}}$

=> ${x = - 1\pm \sqrt{2\over3}}$

I've taken a wrong turn somewhere, I'm not sure how to get to ${x = 0}$ or ${x = -{4\over3}}$.

Answer 1

As mentioned by other users, the textbook solution is correct:

$$12x^{2}+24x-8x=0$$

$$4\Rightarrow 12x^{2}+16x=0$$

Factoring out $4$ gives us $$4x(3x+4)=0$$

We are left with $2$ equations $$4x=0$$ or $$3x+4=0.$$

and solving for these equations gives us $x=0$ or $x=-\frac{4}{3}$.

Answer 2

No, if you consider $12x^2 + 24x = 8$, then considering $x=0$ you get $0=8$, which is clearly wrong.

Considering $12x^2 + 24x = 8x$ this is the same as $12x^2 + 16x = 0$, and from here:

$$ x(12x + 16) = 0$$

Which is fulfilled if $x=0$ or $12 x = -16 \implies x = - \frac{4}{3}$

Answer 3

$12x^2+24x-8x=0\Rightarrow 12x^2+16x=0 \Rightarrow 3x^2+4x=0 \Rightarrow x(3x+4)=0$

So, $x=0$ or $x=- \frac 43$