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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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    $\begingroup$ Did you intend to write $\,x_1 = 9/3\,$ (vs. $3/9$ above) as in the guess in irchans's answer, or did you actually use a different erroneous method? $\endgroup$
    – Bill Dubuque
    Oct 18, 2018 at 15:41

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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 $$(x-3)=0\quad\mathrm{or}\quad(x+1)=0.$$

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    $\begingroup$ Good pedagogy. Plus one. $\endgroup$
    – Lubin
    Oct 18, 2018 at 4:26
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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 $x_2=3$.

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    $\begingroup$ This does not answer the question. You show an alternative method to get the answer, but the real question was "I'm curious to know which step is incorrect". $\endgroup$
    – user193810
    Oct 18, 2018 at 8:47
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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.

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    $\begingroup$ @Pakk OK, I have edited to include all this information. $\endgroup$
    – Parcly Taxel
    Oct 18, 2018 at 8:58
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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 $$3x^2 - 6x^2 - 9 = 0$$ $$x^2 - 2x^2 -3 = 0$$ $$x^2 - 2x^2 +1 = 4$$ $$(x-1)^2 = 2^2$$ $$x-1=2 \text{ or } x-1=-2$$ $$x=3 \text{ or } x=-1$$

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    $\begingroup$ This does not answer the question. You show a correct alternative method to get the answer, but the real question was "I'm curious to know which step is incorrect". $\endgroup$
    – user193810
    Oct 18, 2018 at 8:47
  • $\begingroup$ @Pakk This part has already been answered by irchans and Parcly Taxel, so there is no need for Henry to say the same thing. It still can be useful to other users to know the correct way to approach this problem. $\endgroup$
    – Toby Mak
    Oct 18, 2018 at 12:20
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    $\begingroup$ @TobyMak: Following that logic: the correct way to approach this problem was already given by irchans, so there is no need for Henry to say the same thing. $\endgroup$
    – user193810
    Oct 18, 2018 at 13:44

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