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Question Number 27975 by NECx last updated on 18/Jan/18

$$solvetheinequality$$ $$\frac{1}{x^2+x+1}>0$$

Answered by mrW2 last updated on 18/Jan/18

$$\frac{1}{x^2+x+1}=\frac{1}{x^2+2\times \frac{1}{2}x+{\left(\frac{1}{2}\right)}^2+\frac{3}{4}}=\frac{1}{{(x+\frac{1}{2})}^2+\frac{3}{4}}>0$$ $$\Rightarrow -\mathrm{\infty }<x<+\mathrm{\infty }$$

Commented byRasheed.Sindhi last updated on 18/Jan/18

N̸i̸c̸e̸ S̸i̸r̸!̸

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