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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?

PrincessEev
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Shooting Stars
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  • When you graphed $y=x^2+4x+3$ did you notice which points of the graph were above the $x$-axis. Which intervals of the $x$-axis did those points correspond to? – John Wayland Bales Oct 22 '21 at 03:47
  • For the interval: $(-3,-1)$, the function was below the $x$-axis – Shooting Stars Oct 22 '21 at 03:50
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    If a product such as $(x+1)(x+3)$ is positive, both factors are positive or both factors are negative – J. W. Tanner Oct 22 '21 at 03:53
  • Does this answer your question? quadratic inequality. You can complete the square of any quadratic and then use algebra to determine the signs of the inequalities. You can check the solution by sketching $(x+a)^2 - b > 0$ (for example, $(x+1)^2 - 4 > 0$). – Toby Mak Oct 22 '21 at 06:22

3 Answers3

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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 $x<-3$.

PrincessEev
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Option:

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

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

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

Peter Szilas
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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:

enter image description here

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.

Toby Mak
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  • That's also a good explanation, which I resorted to in the end. However, I wanted to solve this algebraically(without using graphs) – Shooting Stars Oct 22 '21 at 06:23