Quadratic Equations
Quadratic inequalities and geometry
Grade 11
Question:
<p>72. If \(ax^2 + bx + c = 0\) has imaginary roots and \(a - b + c > 0\), then the set of points \((x, y)\) satisfying the equation \(\left|a\left(x^2 + \dfrac{y}{a}\right) + (b+1)x + c\right| = |ax^2 + bx + c| + |x + y|\) consists of the region in the \(xy\)-plane which is</p>
<p>(1) on or above the bisector of I and III quadrant</p>
<p>(2) on or above the bisector of II and IV quadrant</p>
<p>(3) on or below the bisector of I and III quadrant</p>
<p>(4) on or below the bisector of II and IV quadrant</p>
Step-by-Step Solution
Key Concept: Since ax² + bx + c = 0 has imaginary roots and a - b + c > 0, we know a > 0 and ax² + bx + c > 0 for all real x. The absolute value equation simplifies using this positivity to extract constraints on the region.
<p><strong>Step 1: Analyze the given quadratic.</strong><br>Since ax² + bx + c = 0 has imaginary roots, its discriminant b² - 4ac < 0. Combined with a - b + c > 0, this means f(1) = a - b + c > 0. Since the discriminant is negative, ax² + bx + c maintains the sign of a, so a > 0. Therefore, <strong>ax² + bx + c > 0 for all real x</strong>.</p><p><strong>Step 2: Simplify the absolute value equation.</strong><br>The equation is |a(x² + y/a) + (b+1)x + c| = |ax² + bx + c| + |x + y|<br>Rewrite LHS: |ax² + bx + c + x + y| = |ax² + bx + c| + |x + y|<br>Since ax² + bx + c > 0 always, this becomes: |ax² + bx + c + x + y| = (ax² + bx + c) + |x + y|</p><p><strong>Step 3: Apply absolute value inequality property.</strong><br>For the equation |A + B| = A + |B| to hold (where A = ax² + bx + c > 0):<br>We need (A + B) ≥ 0 and either:<br>• B ≥ 0, or<br>• A + B ≥ 0 and -A - B = A + |B| (impossible since A > 0)<br>Therefore, <strong>x + y ≥ 0</strong>.</p><p><strong>Step 4: Verify the constraint.</strong><br>When x + y ≥ 0: |ax² + bx + c + x + y| = ax² + bx + c + x + y<br>And: (ax² + bx + c) + |x + y| = ax² + bx + c + x + y ✓<br>This confirms the region is <strong>x + y ≥ 0</strong>, or equivalently <strong>y ≥ -x</strong>.</p><p>∴ Answer: The region above/on the line y = -x (or half-plane where x + y ≥ 0)</p>
Correct Answer: 3