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Solve for \(x\) in the equation \(x^{2}+5 x+6 \geq 0\)
in Mathematics by Diamond (40,709 points)
reopened by | 35 views

1 Answer

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Answer

\(x \leq-3, x \geq-2\)

 

Explanation

(1) Factor \(x^{2}+5 x+6\)
\[
(x+2)(x+3) \geq 0
\]
(2) Solve for \(x\).
\[
x=-2,-3
\]
(3) From the values of \(x\) above, we have these 3 intervals to test.
\[
\begin{aligned}
&x \leq-3 \\
&-3 \leq x \leq-2 \\
&x \geq-2
\end{aligned}
\]

(4) Pick a test point for each interval.
For the interval \(x \leq-3\)
Let's pick \(x=-4\). Then, \((-4)^{2}+5 \times-4+6 \geq 0\).
After simplifying, we get \(2 \geq 0\), which is true.
Keep this interval..
For the interval \(-3 \leq x \leq-2\) :
Let's pick \(x=-\frac{5}{2}\). Then, \(\left(-\frac{5}{2}\right)^{2}+5 \times \frac{-5}{2}+6 \geq 0\).
After simplifying, we get \(-0.25 \geq 0\), which is false.
Drop this interval..
For the interval \(x \geq-2\) :
Let's pick \(x=0\). Then, \(0^{2}+5 \times 0+6 \geq 0\).
After simplifying, we get \(6 \geq 0\), which is true.
Keep this interval..
(5) Therefore,
\(x \leq-3, x \geq-2\)

 

by Platinum (122,714 points)

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