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# Algebra

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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

Oct 24, 2022

#1
+118132
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$$(x^2 + 2x + 5 - x - 4)(3x^2 - x - 4 + 2x + 8) \ge 0\\ (x^2 + x + 1)(3x^2 + x +4) \ge 0\\ \qquad \text{consider the roots of}\; x^2+x+1\\ \qquad \triangle=b^2-4ac=1-4=-3<0 \qquad\text{ no real roots}\\ \qquad \text{consider the roots of}\; 3x^2+x+4\\ \qquad \triangle=b^2-4ac=1-48=-47<0 \qquad\text{ no real roots}\\$$

Since the coefficiant of x^4 is positive, and there are not real roots.

this funtion will always be positve.

x is in the set of real numbers (no restrictions)

Oct 24, 2022