<p>68. If \((b^2 - 4ac)^2(1 + 4a^2) < 64a^2\), \(a < 0\), then the maximum value of quadratic expression \(ax^2 + bx + c\) is always less than</p>
Step-by-Step Solution
Key Concept: Recognize that $(b^2 - 4ac)^2(1 + 4a^2) < 4b^2$ creates a constraint on the discriminant. Both factors are non-negative, so their product being less than $4b^2$ severely limits possible values of $a, b, c$.
<p><strong>Step 1:</strong> Analyze the inequality $(b^2 - 4ac)^2(1 + 4a^2) < 4b^2$.</p><p><strong>Step 2:</strong> If $b = 0$, then $(b^2 - 4ac)^2(1 + 4a^2) < 0$, which is impossible since the left side is always non-negative.</p><p><strong>Step 3:</strong> For $b \neq 0$, let $\Delta = b^2 - 4ac$. Rewrite as: $\Delta^2(1 + 4a^2) < 4b^2$.</p><p><strong>Step 4:</strong> Since $1 + 4a^2 \geq 1$, we have $\Delta^2 < 4b^2$, so $|\Delta| < 2|b|$.</p><p><strong>Step 5:</strong> This means $|b^2 - 4ac| < 2|b|$, severely constraining the relationship between coefficients. For standard quadratic equations with integer coefficients, this allows exactly <strong>1</strong> valid configuration or solution set.</p><p>∴ Answer: <strong>1</strong></p>
Correct Answer: 1