Quadratic Equations
Roots and their properties
Grade 11
Question:
<p>If \((m_r, 1/m_r)\), \(r = 1, 2, 3, 4\), are four pairs of values of \(x\) and \(y\) that satisfy the equation \(x^2 + y^2 + 2yx + 2fy + c = 0\), then the value of \(m_1 \cdot m_2 \cdot m_3 \cdot m_4\) is</p>
<p>(1) 0</p>
<p>(2) 1</p>
<p>(3) -1</p>
<p>(4) none of these</p>
Step-by-Step Solution
Key Concept: Since (m_r, 1/m_r) satisfies the given equation for four values of r, substitute x = m_r and y = 1/m_r to obtain a quartic equation in m_r. The product of all four roots equals the constant term divided by the leading coefficient.
<p><strong>Step 1:</strong> Substitute (m_r, 1/m_r) into the equation x² + y² + 2yx + 2fy + c = 0:</p><p>m_r² + (1/m_r)² + 2(1/m_r)·m_r + 2f(1/m_r) + c = 0</p><p><strong>Step 2:</strong> Simplify the equation:</p><p>m_r² + 1/m_r² + 2 + 2f/m_r + c = 0</p><p><strong>Step 3:</strong> Multiply through by m_r² to eliminate denominators:</p><p>m_r⁴ + 1 + 2m_r² + 2fm_r + cm_r² = 0</p><p>m_r⁴ + (2 + c)m_r² + 2fm_r + 1 = 0</p><p><strong>Step 4:</strong> This is a quartic equation with four roots m₁, m₂, m₃, m₄. By Vieta's formulas, the product of all roots equals the constant term divided by the leading coefficient:</p><p>m₁·m₂·m₃·m₄ = 1/1 = 1</p><p>∴ Answer: <strong>B</strong> (which is 1)</p>
Correct Answer: B