Prove that the equation x^2 + ax + a - 2 = 0 has two distinct real roots for every real value of a.

Step-by-step solution

1. 1

   Recall what controls the number of real roots. For $Ax^2+Bx+C=0$ with $A\neq 0$, the discriminant

   $\Delta=B^{2}-4AC$

   decides everything: $\Delta>0$ gives two distinct real roots, $\Delta=0$ a repeated root, $\Delta<0$ no real roots. So the whole proof reduces to showing $\Delta>0$ for every $a$.

2. 2

   Identify the coefficients. Careful here — the letter $a$ is a parameter, not the leading coefficient:

   $A=1,\qquad B=a,\qquad C=a-2.$

3. 3

   Compute the discriminant as a function of a.

   $\Delta=a^{2}-4(1)(a-2)=a^{2}-4a+8.$

4. 4

   Complete the square to expose the sign. Half of $-4$ is $-2$, and $(-2)^2=4$:

   $\Delta=\left(a^{2}-4a+4\right)+4=(a-2)^{2}+4.$

   Expanding back gives $a^2-4a+4+4=a^2-4a+8$, confirming the rewrite.

5. 5

   Draw the conclusion. A real square is never negative, so $(a-2)^2\ge 0$ and therefore

   $\Delta=(a-2)^{2}+4\ge 4>0$

   for every real $a$. The discriminant is strictly positive — in fact bounded away from zero — so the equation always has two distinct real roots.

6. 6

   Note where the minimum sits. The smallest possible discriminant, $\Delta=4$, occurs at $a=2$, where the equation becomes $x^2+2x+0=0$ with roots $x=0$ and $x=-2$: distinct, as promised.