Quadratic Equations
Common Roots
Grade 11
Question:
<p>If the equations <span class="math">\(ax^2 + bx + c = 0\)</span> and <span class="math">\(cx^2 + bx + a = 0\)</span>, where <span class="math">\(a, b, c \in \mathbb{R}\)</span> and <span class="math">\(ac \neq 0\)</span>, have a common non-real root, then</p>
<p>(A) <span class="math">\(a = -c\)</span></p>
<p>(B) <span class="math">\(a = c\)</span></p>
<p>(C) <span class="math">\(|b| > 2|a|\)</span></p>
<p>(D) <span class="math">\(|b| < 2|a|\)</span></p>
Step-by-Step Solution
Key Concept: If α is a common non-real root of both equations, then α satisfies both ax² + bx + c = 0 and cx² + bx + a = 0. Subtracting these equations reveals a relationship between the coefficients. For non-real roots of a quadratic with real coefficients, complex conjugate roots appear in pairs.
<p><strong>Step 1: Set up equations for common root.</strong> Let α be the common non-real root. Then:<br/>aα² + bα + c = 0 ... (1)<br/>cα² + bα + a = 0 ... (2)</p><p><strong>Step 2: Subtract equation (2) from equation (1).</strong><br/>aα² + bα + c - (cα² + bα + a) = 0<br/>aα² - cα² + c - a = 0<br/>(a - c)α² + (c - a) = 0<br/>(a - c)α² - (a - c) = 0<br/>(a - c)(α² - 1) = 0</p><p><strong>Step 3: Analyze the factored result.</strong> Since α is non-real, α² ≠ 1 (because 1 is real). Therefore:<br/>a - c = 0<br/>∴ a = c</p><p><strong>Step 4: Verify this makes sense.</strong> If a = c, the second equation cα² + bα + a = 0 becomes aα² + bα + a = 0. This is NOT identical to the first equation aα² + bα + c = 0 unless a = c. When a = c, both equations become aα² + bα + a = 0, which can have non-real roots. The discriminant is b² - 4a², and for non-real roots we need b² < 4a², meaning |b| < 2|a|.</p><p><strong>Step 5: Conclude.</strong> The necessary condition for a common non-real root to exist is a = c.</p><p><strong>∴ Answer:</strong> B</p>
Correct Answer: B