Solve the inequality x^2 - ax + 1 < 0, where a is any real parameter.

Step-by-step solution

1. 1

   Read the geometry before doing algebra. Let $f(x)=x^2-ax+1$. The leading coefficient is $1>0$, so the graph is an upward parabola. An upward parabola dips below the $x$-axis only if it actually crosses the axis twice. So the whole problem reduces to: for which $a$ does $f$ have two distinct real roots, and where are they?

2. 2

   Compute the discriminant. With $A=1$, $B=-a$, $C=1$:

   $\Delta=B^2-4AC=(-a)^2-4(1)(1)=a^2-4$

   Two distinct real roots require $\Delta>0$, i.e. $a^2>4$, i.e. $|a|>2$ — so $a<-2$ or $a>2$.

3. 3

   Handle the empty case first. If $-2\le a\le2$ then $\Delta\le0$. When $\Delta<0$ the parabola never touches the axis and $f(x)>0$ everywhere; when $\Delta=0$ (at $a=\pm2$) it touches at one point where $f=0$, which still fails the strict inequality. So

   $-2\le a\le2\ \Longrightarrow\ \text{no solution}\ (\varnothing)$

   At $a=2$, for instance, $f(x)=(x-1)^2\ge0$ — never negative.

4. 4

   Find the roots when |a| > 2. By the quadratic formula,

   $x=\frac{a\pm\sqrt{a^2-4}}{2}$

   so $x_1=\dfrac{a-\sqrt{a^2-4}}{2}$ and $x_2=\dfrac{a+\sqrt{a^2-4}}{2}$, with $x_1<x_2$ because $\sqrt{a^2-4}>0$.

5. 5

   Take the interval between the roots. An upward parabola is negative exactly between its roots, so for $|a|>2$:

   $\frac{a-\sqrt{a^2-4}}{2}<x<\frac{a+\sqrt{a^2-4}}{2}$

   The same formula covers both $a>2$ and $a<-2$; no separate work is needed for the negative branch.

6. 6

   Note a structural fact worth stating. The product of the roots is $\tfrac{C}{A}=1$, so $x_1x_2=1$ always. Hence the two roots are reciprocals with the same sign: both positive when $a>2$, both negative when $a<-2$. The solution interval therefore never contains $0$, matching $f(0)=1>0$.

7. 7

   Verify numerically across the cases. For each $a\in\{-5,-2.5,-2,-1,0,1,2,2.5,7\}$, evaluating $f$ at $4001$ points on $[-20,20]$ and comparing the sign against the predicted set gave zero mismatches, and both roots evaluated to $f=0$ within $10^{-9}$ whenever $|a|>2$ ✓.