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Find all real values of  x such that
(x^2 + 2x + 5 - x - 4)(3x^2 - x - 4 + 2x + 8) >= 0

 Sep 2, 2022
 #1
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Rewritten in latex, this is \((x^2 + 2x + 5 - x - 4)(3x^2 - x - 4 + 2x + 8) \ge 0\).

 

Let's simplify the things in the parenthases. Doing so, we get \((x^2+x+1)(3x^2+x+4)\ge0.\) These cannot be simplified further.

 

However, all the signs in this are positive. So for any value of \(x\) inputted in here, it's always \(\ge 0\). So the solutions to this are \(-\infty , or in interval,  \(x\in(-\infty,+\infty)\).

 Sep 3, 2022

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