Solving the inequality $x^2 + 4x + 3 > 0$

Question

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?

Answer 1

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 $x<-3$.

Answer 2

Option:

$(x+2)^2>1;$

$\Rightarrow:$ $|x+2|>1$, i. e.

$x+2>1$, or $x+2<-1.$

Answer 3

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.