Solving the inequality $x^2 + 4x + 3 > 0$
So here is the question:
For what values of $x$ is $x^2 + 4x + 3 > 0$?
So I decided to factor the right-hand side into:
$$(x+1)(x+3) > 0$$
getting:
$$x+1>0 \text{ or } x+3 > 0 \implies x > -1 \text{ or } x > -3$$
But this is incorrect because I graphed the function and I saw $x\not>$ $-3$ for some values of $x$ as the graph dips below the $x$-axis. Where I am going wrong? Could someone help?
$\endgroup$ 43 Answers
$\begingroup$You have to consider several scenarios.
In general, when is $ab>0$? This happens when
- $a,b$ are both positive, or
- $a,b$ are both negative
So, consider: you have $(x+1)(x+3) > 0$. Then,
- $x+1,x+3$ are both positive, or
- $x+1,x+3$ are both negative
In your working, you assume only the former, and get that either $x > -1$ or $x>-3$. Of course, if both are positive, we need $x>-1$, so that gives one solution.
But what if both are negative? Then, when you solve $(x+1)(x+3) > 0$, you are dividing by one of those factors and therefore must reverse the inequality sign. Hence,
$$(x+1)(x+3) > 0 \implies x+1 < 0 \text{ or } x+3 < 0$$
when both are negative. Therefore, $x < -1$ or $x < -3$. Of course, again, we see that both factors are only negative if $x < -3$.
Thus, the solution set is the set of $x$ where $x > -1$ or $x<-3$.
$\endgroup$ 1 $\begingroup$Option:
$(x+2)^2>1;$
$\Rightarrow:$ $|x+2|>1$, i. e.
$x+2>1$, or $x+2<-1.$
$\endgroup$ $\begingroup$Since the leading term of the quadratic (of $x^2)$ is positive, it must be concave up, not down.
We can draw a quick sketch:
and hence $(x+1)(x+3) > 0$ when $x < -3, x > - 1$. When $x = -1$ or $-3$, you get $0 > 0$ which is not true.
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