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How can I show that $x^2 +x +1$ is aways positive for all values of $x$? Do I use discriminant or completing the square?

Pedro
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user279584
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    Either will do. I prefer completing the square. And because of an allergy to fractions, I prefer to say that $x^2+x+1$ is always positive if and only if $4x^2+4x+4$ is always positive, and $4x^2+4x+4=(2x+1)^2+3$. – André Nicolas Oct 12 '15 at 16:30
  • Complete the square to get $x^2+x+1=(x+\frac{1}{2})^2+\frac{3}{4}$. – ThePuix Oct 12 '15 at 16:31

1 Answers1

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$x^2+x+1$

$=(x^2+2x+1)-x$

$=(x+1)^2-x$

Now,$(x+1)$ $>$ $x$.

So,$(x+1)^2$ $>$ $x$

So,$(x+1)^2$ $-$ $x$ is positive.

Soham
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