Quadratic Equations
Vieta's Relations
Grade 11
Question:
<p>Let <i>r</i><sub>1</sub>, <i>r</i><sub>2</sub>, <i>r</i><sub>3</sub>, ..., <i>r</i><sub><i>n</i></sub> be <i>n</i> positive integers, not necessarily distinct, such that <br/><i>(x</i> − <i>r</i><sub>1</sub>)(<i>x</i> − <i>r</i><sub>2</sub>)···(<i>x</i> − <i>r</i><sub><i>n</i></sub>) = <i>x</i><sup><i>n</i></sup> − 56<i>x</i><sup><i>n</i>−1</sup> + ... − 2009.</p><p>The possible value of <i>n</i> is</p>
<p>(A) 4</p>
<p>(B) 5</p>
<p>(C) 8</p>
<p>(D) 9</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to relate the sum and product of roots to coefficients, then factor the product to find the possible roots.
<p><strong>Step 1:</strong> By Vieta's formulas, the sum of roots is <i>r</i><sub>1</sub> + <i>r</i><sub>2</sub> + ... + <i>r</i><sub><i>n</i></sub> = 56, and the product is <i>r</i><sub>1</sub>·<i>r</i><sub>2</sub>···<i>r</i><sub><i>n</i></sub> = (−1)<sup><i>n</i></sup>·(−2009) = (−1)<sup><i>n</i>+1</sup>·2009.</p><p><strong>Step 2:</strong> Since 2009 = 7² × 41, we need <i>r</i><sub>1</sub>·<i>r</i><sub>2</sub>···<i>r</i><sub><i>n</i></sub> = 2009 (assuming <i>n</i> is odd).</p><p><strong>Step 3:</strong> We need positive integer roots whose product is 2009 and sum is 56. No single root can be ≥ 49 (as the sum would exceed 56). We find: 56 = 41 + 7 + 7 + 1, so <i>r</i><sub>1</sub> = 41, <i>r</i><sub>2</sub> = <i>r</i><sub>3</sub> = 7, <i>r</i><sub>4</sub> = 1, and 41 × 7 × 7 × 1 = 2009.</p><p><strong>Step 4:</strong> Therefore, <i>n</i> = 4.</p><p>∴ Answer is (A).</p>
Correct Answer: A