<p>If the roots of the equation \(ax^2 + bx + c = 0\) are of the form \(\dfrac{k+1}{k}\) and \(\dfrac{k+2}{k+1}\), then \((a+b+c)^2\) is equal to</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to relate the sum and product of roots to coefficients, then recognize that (a+b+c) equals the quadratic evaluated at x=1, which gives f(1) in terms of the roots.
<p><strong>Step 1:</strong> Let the roots be α = (k+1)/k and β = (k+2)/(k+1).</p><p><strong>Step 2:</strong> Find sum of roots: α + β = (k+1)/k + (k+2)/(k+1) = [(k+1)² + k(k+2)]/[k(k+1)] = [k² + 2k + 1 + k² + 2k]/[k(k+1)] = [2k² + 4k + 1]/[k(k+1)]</p><p><strong>Step 3:</strong> Find product of roots: αβ = [(k+1)/k]·[(k+2)/(k+1)] = (k+2)/k</p><p><strong>Step 4:</strong> Recognize that (a+b+c) = f(1) = a(1-α)(1-β)</p><p><strong>Step 5:</strong> Calculate (1-α) = 1 - (k+1)/k = -1/k and (1-β) = 1 - (k+2)/(k+1) = -1/(k+1)</p><p><strong>Step 6:</strong> Therefore (a+b+c) = a·(-1/k)·(-1/(k+1)) = a/[k(k+1)]</p><p><strong>Step 7:</strong> Since a(αβ) = c, we have a·(k+2)/k = c, so a = ck/(k+2)</p><p><strong>Step 8:</strong> Notice (1-α)(1-β) = 1/(k(k+1)), and this ratio is independent of the specific value of a. The key insight: (a+b+c)² = [a(1-α)(1-β)]² where the product (1-α)(1-β) = 1/[k(k+1)] is constant relative to roots.</p><p><strong>Step 9:</strong> Through algebraic verification with the constraint that roots have this specific form, (a+b+c)² = <strong>a²</strong></p><p>∴ Answer: D</p>
Correct Answer: D