mjhjmtu

Very Easy question, but I wasn't actively doing the question and I got it wrong. I was looking at it & didn't know which step was wrong? It looked like all the individual steps are correct but the final solution is wrong.

I'm fully aware the solution is wrong & that if I plug the "roots" back in its wrong.

$$ 3x^2 - 6x^2 - 9 = 0 $$
$$ 9 = 3x(x-2) $$
$$ x_1 = frac39 qquad x_2 = 11 $$

I should have factored etc etc, but I'm curious to know which step is incorrect & why.

I think you meant to write $3x^2−6x−9=0$. Then your first steps are fine:

$$3x^2−6x=9$$
$$3x(x-2)=9$$

At this point, I think you erroneously separated this equation into two equations
$$3x=9quadmathrmor$$
$$x-2=9$$.

If you have two numbers $a$ and $b$ where $acdot b=9$, you cannot conclude that $a=9$ or $b=9$.

On the other hand, (*) if you have two numbers $a$ and $b$ where $acdot b=0$, you can conclude that $a=0$ or $b=0$.

Let's do it right.
$$3x^2−6x−9=0$$
$$x^2−2x−3=0$$
$$(x-3)(x+1)=0$$

Now apply (*), to get
$$(x-3)=0quadmathrmorquad(x+1)=0.$$

You cannot have a non-zero number on the RHS of $(x-a)(x-b)=k$ (in this case, it is 9). Only when $(x-a)(x-b)=0$ can wd conclude that $x=a$ and $x=b$ are roots.

A root of a function is a value $x$ such that $f(x)=0$. However, you have an expression $g(x)=9$, so we are not finding roots. In this case, $f(x)=3x^2-6x-9$ so to find the roots is to solve
$$3x^2-6x-9=0$$
Factoring, we obtain
beginalign*
3(x^2-2x-3) & =0
\ 3(x+1)(x-3) & = 0
\ (x+1)(x-3) & =0
endalign*
In order for the equality to hold, either $x+1=0$ or $x-3=0$, so the roots are $x_1=-1$ and $x_2=3$.