If $a$ and $b$ are integers such that $2x^2 - ax + 2 > 0$ and $x^2 −bx + 8 \geq 0$ for all real numbers $x,$ then the largest possible value of $2a-6b$ is

We have two conditions that must hold for all real numbers x:

$2x^2 - ax + 2 > 0$ (always positive)

$x^2 - bx + 8 \geq 0$ (always non-negative)

Key Insight: When a quadratic expression is always positive (or non-negative) for all real values of x, we can use the discriminant to find constraints on the coefficients.

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For $2x^2 - ax + 2 > 0$ to be always positive, the parabola must never touch or cross the x-axis.

If a quadratic $Ax^2 + Bx + C$ has discriminant $B^2 - 4AC < 0$, then it has no real roots. Since the coefficient of $x^2$ is positive (2 > 0), the parabola opens upward. No real roots + upward opening = always positive.

For $2x^2 - ax + 2$:

$A = 2$, $B = -a$, $C = 2$

Discriminant: $(-a)^2 - 4(2)(2) < 0$

$a^2 - 16 < 0$

$a^2 < 16$

$-4 < a < 4$

Since $a$ is an integer: $a \in \{-3, -2, -1, 0, 1, 2, 3\}$

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For $x^2 - bx + 8 \geq 0$ to be always non-negative, the parabola must never go below the x-axis.

We use $\leq 0$ for the discriminant here because we allow the parabola to touch the x-axis (that's why we have $\geq 0$ instead of $> 0$).

For $x^2 - bx + 8$:

$A = 1$, $B = -b$, $C = 8$

Discriminant: $(-b)^2 - 4(1)(8) \leq 0$

$b^2 - 32 \leq 0$

$b^2 \leq 32$

$-\sqrt{32} \leq b \leq \sqrt{32}$

$-4\sqrt{2} \leq b \leq 4\sqrt{2}$

Since $\sqrt{2} \approx 1.414$, we have $4\sqrt{2} \approx 5.66$

Since $b$ is an integer: $b \in \{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\}$

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To maximize $2a - 6b$, we need:

Largest possible value of $a$: From our constraint, $a_{max} = 3$

Smallest possible value of $b$: From our constraint, $b_{min} = -5$

Since we're subtracting $6b$, subtracting a negative number gives us a larger result.

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

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Answer: 36