<p>If a, b, c, d are four consecutive terms of an increasing AP, then the roots of the equation \((x-a)(x-c) + 2(x-b)(x-d) = 0\) are</p>
<p>(a) real and distinct</p>
<p>(b) non-real complex</p>
<p>(c) real and equal</p>
<p>(d) integers</p>
Step-by-Step Solution
Key Concept: Use the property that consecutive AP terms can be written as a-3r, a-r, a+r, a+3r (where r is common difference), then substitute to reveal that the quadratic factors into a perfect square or has a special structure.
<p><strong>Step 1:</strong> Let the four consecutive AP terms be a, b, c, d with common difference r.<br/>So: a = a, b = a+r, c = a+2r, d = a+3r</p><p><strong>Step 2:</strong> Alternatively, use symmetric notation: Let a = m-3r, b = m-r, c = m+r, d = m+3r (centered at m)<br/>Then: x-a = x-m+3r, x-b = x-m+r, x-c = x-m-r, x-d = x-m-3r</p><p><strong>Step 3:</strong> Substitute into the equation:<br/>(x-m+3r)(x-m-r) + 2(x-m+r)(x-m-3r) = 0<br/>= [(x-m)+3r][(x-m)-r] + 2[(x-m)+r][(x-m)-3r]</p><p><strong>Step 4:</strong> Let y = x-m:<br/>(y+3r)(y-r) + 2(y+r)(y-3r) = 0<br/>= y² + 2yr - 3r² + 2(y² - 2yr - 3r²) = 0<br/>= y² + 2yr - 3r² + 2y² - 4yr - 6r² = 0<br/>= 3y² - 2yr - 9r² = 0</p><p><strong>Step 5:</strong> Using the quadratic formula or factoring:<br/>3y² - 2yr - 9r² = 0<br/>This factors or gives y = (2r ± √(4r² + 108r²))/(6) = (2r ± √(112r²))/6 = (2r ± 4r√7)/6<br/><br/>More directly: the roots simplify to real, distinct values lying symmetrically about the AP's center.</p><p><strong>Conclusion:</strong> The roots are real and distinct (lying between and outside the given four terms depending on the discriminant sign).</p><p>∴ Answer: A</p>
Correct Answer: A