Quadratic Equations
Roots of cubic equations
Grade 11
Question:
<p>If <i>a</i>, <i>b</i>, <i>c</i>, <i>p</i>, <i>q</i>, <i>r</i> are non-zero real numbers, such that <i>a</i> < <i>b</i> < <i>c</i> and</p><p>\[f(x) = (x - a)(x - b)(x - c) - p^2(x - a) - q^2(x - b) - r^2(x - c),\]</p><p>then \(f(x) = 0\) must have</p>
<p>(A) exactly 1 real root</p>
<p>(B) exactly 3 distinct real roots</p>
<p>(C) 2 equal and 1 distinct real root</p>
<p>(D) nothing can be said</p>
Step-by-Step Solution
Key Concept: Use the Intermediate Value Theorem with sign changes of a cubic function at strategically chosen points to guarantee three distinct real roots.
<p><strong>Analysis:</strong> Consider \(f(x) = (x-a)(x-b)(x-c) - p^2(x-a) - q^2(x-b) - r^2(x-c)\). Rewrite as:</p><p>\[f(x) = (x-a)[(x-b)(x-c) - p^2] - q^2(x-b) - r^2(x-c)\]</p><p>At \(x = a\): \(f(a) = -q^2(a-b) - r^2(a-c)\). Since \(a < b < c\) and \(q^2, r^2 > 0\), we have \(f(a) > 0\).</p><p>At \(x = b\): \(f(b) = (b-a)(-p^2) - r^2(b-c) = -p^2(b-a) - r^2(b-c)\). Since \(b-a > 0\) and \(b-c < 0\), \(f(b) < 0\).</p><p>At \(x = c\): \(f(c) = (c-a)(c-b) - p^2(c-a) - q^2(c-b) > 0\) (by similar analysis).</p><p>Since \(f\) is a cubic with positive leading coefficient, and \(f(a) > 0\), \(f(b) < 0\), \(f(c) > 0\), by Intermediate Value Theorem, \(f\) has exactly 3 distinct real roots.</p><p>∴ Answer is <strong>B</strong>.</p>
Correct Answer: B