Largest possible value of linear expression

For $2x^2 - ax + 2 > 0$ to hold for all real $x$, its discriminant must be strictly negative: $D_1 = a^2 - 4(2)(2) < 0 \implies a^2 < 16$ Since $a$ is an integer, $-3 \le a \le 3$.

For $x^2 - bx + 8 \ge 0$ to hold for all real $x$, its discriminant must be non-positive: $D_2 = b^2 - 4(1)(8) \le 0 \implies b^2 \le 32$ Since $b$ is an integer, $-5 \le b \le 5$.

To maximize $2a - 6b$:

• We choose the maximum possible value for $a$: $a = 3$.
• We choose the minimum possible value for $b$: $b = -5$.

Largest possible value of $2a - 6b = 2(3) - 6(-5) = 6 + 30 = 36$.