# A doubt in quadratic inequality

While I was doing a problem I came upon this: $$(a^2-4)>0$$ $$(a-2)(a+2)>0$$

Now I thought it will be

$a>2$ or $a>-2$

but it was $a<-2$ or $a>2$

Can u explain me why it is so?

Thank you

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An alternative to the answers given already: sketch the graph of $y=(x-2)(x+2)$. When are the $y$-values positive? – Kelvin Soh Dec 25 '13 at 7:09
$bc>0$ means that $b,c$ are not zero and have the same sign. In this case, that means that either (1) $a-2>0$ and $a+2>0$, or (2) $a-2<0$ and $a+2<0$. Case (1) means that $a>2$ and $a>-2$. Case (2) means that $a<2$ and $a<-2$. Note that case (1) is just the same as $a>2$, and case (2) is just the same as $a<-2$: If $a>2$ then clearly $a>-2$. Similarly, if $a<-2$, then also $a<2$. Now, to see that your conclusion is incorrect, note that, for example, $a=0$ satisfies $a>-2$, so it satisfies the disjunction $a>2$ or $a>-2$. However, $(0-2)(0+2)=-4<0$. – Andrés E. Caicedo Dec 25 '13 at 7:11
Thank you all. I dont know how i missed it. I just split them into two as in equality and put the sign instaed of = – chndn Dec 25 '13 at 7:16
The most voted gets the correct answer tick :D – chndn Dec 25 '13 at 7:21
– Mhenni Benghorbal Dec 25 '13 at 23:45

Now the inequality is valid if * $(a-2)$ and $(a+2)$ are both positive or are both negative. Now $$(a-2)>0 ~\hbox{and}~ (a+2)> 0 \Rightarrow a>2$$ $$(a-2)<0 ~\hbox{and}~ (a+2)< 0 \Rightarrow a<-2$$

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Hint

$$xy>0\iff (x>0\land y>0)\lor(x<0\land y<0)$$

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$$(a+2)(a-2)\gt0$$ means that either "$a+2$ and $a-2$ are positive" or "$a+2$ and $a-2$ are negative", so?

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If $x, y \in \mathbb{R}$ and $xy > 0$, either $x, y > 0$ or $x, y < 0$. As $(a-2)(a+2) > 0$, then either $a - 2 > 0$ and $a + 2 > 0$ which occurs when $a > 2$, or $a - 2 < 0$ and $a + 2 < 0$ which occurs when $a < -2$.

Alternatively, given $a^2 - 4 > 0$ we have $a^2 > 4$ so $\sqrt{a^2} > 2$ because $\sqrt{x}$ is an increasing function. As $|a| = \sqrt{a^2}$ we see that $|a| > 2$ so $a < -2$ or $a > 2$.

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Others have already answered this question, but I'd like to give a hint. If your answer differs from the expected one, find a solution that gives different results for a quick check.

You say a > -2 or a > 2; another answer is a < -2 or a > 2. Where do these solutions differ? Most obviously, when a = 0.

Replace a with 0, and what do you get? 0*0 - 4 = -4 , therefore 0 is not a part of solution, therefore your answer is invalid. When you know that you've made a mistake, you may go on to actually searching for the error. Otherwise you may waste time trying to fix the right answer: textbooks are not immune from errors.

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