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How $x^2 + 5x + 6 \ge 0$ implies $−3 \ge x \ge −2$ ?

I represented $x^2 + 5x + 6 \ge 0$ as $(x+3)(x+2) \ge 0$,which means that either $(x+3) \ge 0$ and $(x+5) \ge 0$ which giving $x \ge -3$ or $x \ge -2$ but how does this gives $−3 \ge x \ge −2$?

ADDED: The above expression could also mean that $(x+3) \le 0$ and $(x+2) \le 0$ which will give $x \le -3$ and $x \le -2$,now should combine this two to get that result? but in that case we would get both $−3 \le x \le −2$ and $−2 \le x \le −3$ (if I am not wrong) but then the first one should be takes as the answer is taken as the second one makes no sense,am I right?

Source of the problem. Check the solution given there which causes the confusion.

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HINT $\ $ You forgot the case that both factors are $\le 0\:.$ – Bill Dubuque Jul 23 '11 at 23:04
@Bill Dubuque:I think I got the answer :) – Quixotic Jul 23 '11 at 23:06
Not quite. First your $x+5$ should be $x+2$. Second the factors should have the same sign. Combine the inequalities from both cases and all will be clear. – Bill Dubuque Jul 23 '11 at 23:09
up vote 10 down vote accepted

$−3 \ge x \ge −2$ doesn't make sense, since $-3<-2$. Further, your correct factorization $(x+3)(x+2) \ge 0$ doesn't imply that either $(x+3) \ge 0$ or $(x+2) \ge 0$ (assuming your $5$ here is a typo), but just the opposite, that either both factors are $\ge0$ or neither is. So the correct inference is that $-3\ge x$ or $x\ge-2$, which I presume must somewhere along the way have gotten truncated to the incorrect $−3 \ge x \ge −2$.

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I remember doing this in high school. The teacher got mad. It still makes sense to me; it is unambiguous. – Emre Jul 24 '11 at 22:02

$x^{2}+5x+6=0$, Solutions are $x=-3,x=-2.$

Since $x^{2}+5x+6=(x+3)(x+2)$, we have three cases:

  1. If $x\geq -2$, then $x+3\geq 1$, $x+2\geq 0$, so $x^{2}+5x+6\geq 0.$
  2. If $x\leq -3$, then $x+3\leq 0$, $x+2\leq -1$, so $x^{2}+5x+6\geq 0.$
  3. If $-3<x<-2$, then $x+3>0$, $x+2<0$, so $x^{2}+5x+6<0.$

Remark: instead of "$x^2 + 5x + 6 \ge 0$ implies $−3 \ge x \ge −2$", we should say that "$x^2 + 5x + 6 \le 0$ is equivalent to $−3 \le x \le −2$". The double inequality $−3 \ge x \ge −2$ is a compact form of $−3 \ge x$ and $x \ge −2$, which is not satisfied by any real $x$.

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I strongly advocate for only using $<$ or $\le$ when solving inequalities. Also, as joriki points out, inequalities are transitive so $-2\le x \le -3$ does not make sense since $-3$ is to the left of $-2$. Note that in my usage, "to the left of" and "less than" are synonyms. Your quadratic expression $y=x^2 + 5x +6$ represents a parabola that holds water, dresses to the left and has intercepts at $x=-2$ and $x=-3$. The parabola is below the $x$-axis when $-3<x<-2$ and above the $x$-axis when $x<-3$ or $-2<x$. You can see this by examining the signs of the linear factors in the three intervals: to the left of $-3$, between $-3$ and $-2$ and to the right of $-2$. To the far left both factors are negative, so the result is positive. Between one factor is negative, the other positive, so the result is negative. To the right, both factors are positive.

If you like you can evaluate the quadratic at $x=-10$, $x=-2.5$ and $x=10$ to verify the signs that I indicated.

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HINT $\ $ Simpler than manipulating inequalities is the following. To determine the intervals where a polynomial $\rm\:f(x) \ge 0\:,\:$ partition the real line into intervals whose endpoints are the roots of $\rm\:f\:.\:$ since $\rm\:f(x)\:$ has constant sign on each such open interval $\rm\:I\:$ (by the Intermediate Value Theorem), to determine if $\rm\:f(x)\ge 0\:$ on $\rm\:I\:$ simply evaluate $\rm\:f\:$ at any convenient sample point of the interval. Only finitely many tests need be performed since a polynomial has no more roots than its degree.

In fact the same idea generalizes to higher dimensions, yielding an algorithm to decide the truth of any first order multivariate statements about the reals - as Tarski showed by his celebrated method of quantifier elimination for real-closed fields. This has an algorithmic form that is essentially a higher-dimensional cylindrical version of the finite interval decomposition for the above one dimensional case. For more google "cylindrical algebraic decomposition" or CAD algorithm.

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