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?


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:


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

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

I am very confused.

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    $\begingroup$ Why are there downvotes? This is easy math for several people but effort has been shown. Downvotes are really unnecessary. $\endgroup$ – Daniel W. Farlow Apr 2 '15 at 20:22
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    $\begingroup$ @crash Agreed, though it's come to be expected, no? $\endgroup$ – Jonathan Hebert Apr 2 '15 at 20:24
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    $\begingroup$ The question is pretty clear too. It asks whether there's a flaw in the chain of reasoning. In fact there is a flaw, so the answer is to show where the flaw is. $\endgroup$ – David K Apr 2 '15 at 20:24
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    $\begingroup$ Now I did not dowbvote and it is not a big deal, but then that the title is "Flaw in Mathematics??" is, well, maybe a bit over the top. $\endgroup$ – quid Apr 2 '15 at 20:29
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    $\begingroup$ To reinforce what quid said, the title "Flaw in Mathematics" is bound to be highly provocative (and negatively received). So a more apropos title would be nice. Nonetheless, kudos for being daring enough to post your question and to ask for feedback. You've certainly gotten quite a bit of it! $\endgroup$ – Daniel W. Farlow Apr 2 '15 at 20:32

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 $ab\neq 0$.

  • $\begingroup$ Great answer.. @Spenser please answer this question also in the easiest possible way...math.stackexchange.com/questions/1216246/… $\endgroup$ – Jai Mahajan Apr 2 '15 at 20:57
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    $\begingroup$ @user166748 You should consider placing a bounty on that question, when possible, if you really want the answer. This isn't the way of going about obtaining one. $\endgroup$ – Daniel W. Farlow Apr 2 '15 at 21:10

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.


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

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    $\begingroup$ It's funny that as a counter-example, you picked the one instance (WLOG) that isn't actually a counter-example. $\endgroup$ – Jonathan Hebert Apr 2 '15 at 20:21
  • $\begingroup$ Much better ;-) $\endgroup$ – Jonathan Hebert Apr 2 '15 at 20:24

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

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


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