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.
$$ 3x^2 - 6x - 9 = 0 $$ $$ 9 = 3x(x-2) $$ $$ x_1 = \frac{3}{9} \qquad x_2 = 11 $$
I should have factored etc etc, but I'm curious to know which step is incorrect & why.
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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 $$3x=9\quad\mathrm{or}$$ $$x-2=9$$.
If you have two numbers $a$ and $b$ where $a\cdot b=9$, you cannot conclude that $a=9$ or $b=9$.
On the other hand, (*) if you have two numbers $a$ and $b$ where $a\cdot b=0$, 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 $$(x-3)=0\quad\mathrm{or}\quad(x+1)=0.$$
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 \begin{align*} 3(x^2-2x-3) & =0 \\ 3(x+1)(x-3) & = 0 \\ (x+1)(x-3) & =0 \end{align*} In order for the equality to hold, either $x+1=0$ or $x-3=0$, so the roots are $x_1=-1$ and $x_2=3$.
You cannot have a non-zero number on the RHS of $(x-a)(x-b)=k$ to conclude that $x-a=k$ or $x-b=k$ – indeed, $x-a$ and $x-b$ might be two numbers that surreptitiously multiply to $k$. For example, suppose $k=6$. Then $x-a$ might be 2 and $x-b$ might be 3.
The reason we can conclude from $(x-a)(x-b)=0$ that $x=a$ and $x=b$ are solutions is that it allows the other factor to vary however we want, as zero multiplied by any number remains zero.
There are at least two approaches to quadratics:
Take everything over to one side, factor and conclude that at least one of the factors is zero: $$3x^2 - 6x^2 - 9 = 0$$ $$3(x-3)(x+1)=0$$ $$x=3 \text{ or } x=-1$$
Complete the square on one side leaving a constant on the other and take square roots $$3x^2 - 6x^2 - 9 = 0$$ $$x^2 - 2x^2 -3 = 0$$ $$x^2 - 2x^2 +1 = 4$$ $$(x-1)^2 = 2^2$$ $$x-1=2 \text{ or } x-1=-2$$ $$x=3 \text{ or } x=-1$$
