Prove $x^2 - x + 1 $ is always positive.

Question

While solving a question, I came up with an inequality : $(1+x)(1-x+x^2)>0$

The book stated -

where $(1-x+x^2)$ is always positive as $D<0$ and $a>0$

I'm not that sure how did it conclude that it will be always positive just with the two results ( $ D < 0 \ \& \ a > 0 $ ) .

I know that the Discriminant of the quadratic equation is negative. But how do we conclude that the quadratic equation (given) will always be positive?

Answer 1

As another answer tells you, you can see that the polynomial is always positive by completing the square.

Another way to see this is to note that the discriminant is negative. The discriminant for a quadratic polynomial $ax^2 + bx + c$ is $D = b^4-4ac$. It is a fact that the graph of $y =ax^2 +bx + c$ crosses (or touches) the $x$-axis is $D \geq 0$. So since for your polynomial $D = (-1)^2 - 4 = -3 < 0$, the graph stays entirely on one side of the $x$-axis. Since $a>0$ you have the graph is above the $x$-axis, that is, the polynomial is always positive.

Answer 2

Note that $$x^2-x+1=\left(x-\frac 12\right)^2+\frac 34.$$

Answer 3

Alternatively, let $f(x) = x^2 - x + 1 \Rightarrow f'(x) = 2x-1 = 0 \iff x = 1/2$, and $1/2$ is the only critical point of $f$. Then $f''(1/2) = 2 > 0$, and this shows $f$ attains an absolute minimum at $x = 1/2$ which is $f(1/2) = 3/4$. This means $f(x) \geq 3/4 > 0 \Rightarrow f(x) > 0, \forall x$.