Quadratic Equations
Nature of roots
Grade 11
Question:
<p>75. If \(a, b, c, d\) are four consecutive terms of an increasing A.P., then the roots of the equation \((x - a)(x - c) + 2(x - b)(x - d) = 0\) are</p>
<p>(1) non-real complex</p>
<p>(2) real and equal</p>
<p>(3) integers</p>
<p>(4) real and distinct</p>
Step-by-Step Solution
Key Concept: Substitute the A.P. terms using common difference to express the equation in terms of fewer variables, then recognize the resulting quadratic has a special structure that reveals its roots directly.
<p><strong>Step 1:</strong> Let the four consecutive A.P. terms be: a, a+k, a+2k, a+3k where k > 0 (increasing A.P.)</p><p>So: a=a, b=a+k, c=a+2k, d=a+3k</p><p><strong>Step 2:</strong> Substitute into the equation (x-a)(x-c) + 2(x-b)(x-d) = 0:</p><p>(x-a)(x-a-2k) + 2(x-a-k)(x-a-3k) = 0</p><p><strong>Step 3:</strong> Let y = x-a-k (shift origin to the middle). Then:</p><p>(y+k)(y-k) + 2(y)(y-2k) = 0</p><p>y² - k² + 2y² - 4ky = 0</p><p>3y² - 4ky - k² = 0</p><p><strong>Step 4:</strong> Factor: (3y+k)(y-k) = 0</p><p>So y = -k/3 or y = k</p><p><strong>Step 5:</strong> Convert back: x = y + a + k</p><p>x = -k/3 + a + k = a + 2k/3 or x = k + a + k = a + 2k</p><p>∴ The roots are <strong>a + 2k/3</strong> and <strong>a + 2k</strong> (or equivalently: a + 2/3(c-a) and c)
Correct Answer: 4