Strict positivity requires a strict discriminant bound

x^2-x+1>0

Solution

(i)

Show that $x^2-x+1>0$ for every real $x$ .

Hint

Answer · (i)

x^2-x+1=(x-1/2)^2+3/4>0

(ii)

Hence find all values of $a$ such that $\dfrac{x^2+ax-2}{x^2-x+1}<2$ for every real $x$ .

Hint

Move 2 to the left and combine over $x^2-x+1$ .
The denominator is always positive, so only the numerator matters.
'For every $x$ ' means the numerator quadratic is always negative: negative leading coefficient and negative discriminant.

Answer · (ii)

-6<a<2

Explore the idea

Each step shows q(x) = −x² + (2 + a)x − 4 for a different a. The curve is entirely below the x-axis exactly for −6 < a < 2.

Start: a = −7

$a=-7$ is outside the range: the discriminant $(2+a)^2-16=9>0$ , so $q$ is positive between two roots.

Marked points at this step, rounded to 6 significant figures.
Point	x	y
maximum above 0	-2.5	2.25

Ideas to try

Find the largest and smallest values of $a$ for which the curve never touches the axis.
Why does a larger $|a+2|$ push the peak up?