Why are there four solutions to $x^2-2x-8=0$ in $\mathbb{R}$? Or am I wrong?

Question

It might be a very trivial question to ask but why do we get four different solutions for a quadratic equation using these two methods?

$x^2-2x-8=0$

We see that factors are $(x-4)$ and $(x+2)$ so we get $x=4$ or $- 2$.

Now when we factorise in the following way we get different answers:

$x^2-2x=8$

$x(x-2)=8$ [How can this be!?]

And we get $x=8$ or $x=10$ [How!?]

I am very confused.

Answer 1

Your mistake is to assume that for $a,b,c\in\mathbb{R}$, $$ab=c\quad\implies\quad a=c\quad\text{or}\quad b=c.\tag{1}$$ This is not true in general.

For example $2\times\frac{1}{2}=1$, but neither $2=1$ nor $\frac{1}{2}=1$.

Further explanations: However, $(1)$ is true when $c=0$, and this is why when finding the roots of a polynomial we can factor it and use the argument $$(x-\alpha)(x-\beta)=0\quad\implies\quad x=\alpha\quad\text{or}\quad x=\beta.$$ To prove that if $ab=0$ then $a=0$ or $b=0$, you can argue by contradiction. If both $a\neq 0$ and $b\neq 0$, then surely $ab\neq 0$.

Answer 2

You should notice that's only when ab=0 then a=0 or b=0 but when you have a*b=8 you can't say a=8 or b=8 for instance a could be equal to 4 and b to 2

Answer 3

Spenser's answer is great because it addresses the general issue. Perhaps to add some clarity, let's look at your particular quandary.

You realized that $$ x^2-2x-8 = 0\Longleftrightarrow (x-4)(x+2)=0\Longleftrightarrow x=4,-2. $$ Okay. That's great. Now we can rearrange, as you did, for the following: $$ x^2-2x-8=0\Longleftrightarrow x^2-2x=8\Longleftrightarrow x(x-2)=8\Longleftrightarrow x=? $$ Try plugging in $x=8$ or $x=10$. They do not work. However, plugging in $x=4$ and $x=-2$ works. It's simply less obvious that these $x$-values work when you express your quadratic in this way.

Answer 4

You should do a simple sanity check in a case like this. Plug your answer back into the original equation and make sure it works. Use a calculator and nothing more complicated:

Your tentative answers $x = 4, -2, 8, 10$ check out as

$$(4)^2-2\times(4)-8\stackrel{?}{=}0\space \space \checkmark$$ $$(-2)^2-2\times(-2)-8\stackrel{?}{=}0\space \space \checkmark$$ $$(8)^2-2\times(8)-8\stackrel{?}{=}0\space \space NO!$$ $$(10)^2-2\times(10)-8\stackrel{?}{=}0\space \space NO!$$