<p>If the roots of the equation \(a(b-c)x^2 + b(c-a)x + c(a-b) = 0\) are equal, show that \(2/b = 1/a + 1/c\).</p>
Step-by-Step Solution
Key Concept: For equal roots, the discriminant must be zero. Recognize that the coefficients have a cyclic structure: a(b-c), b(c-a), c(a-b), which sum to zero, making x=1 always a root.
<p><strong>Step 1:</strong> Verify x=1 is always a root. Substitute x=1: a(b-c)(1)² + b(c-a)(1) + c(a-b) = a(b-c) + b(c-a) + c(a-b) = ab - ac + bc - ab + ca - cb = 0 ✓</p><p><strong>Step 2:</strong> Since x=1 is a root and roots are equal, both roots equal 1. By Vieta's formula, product of roots = (constant term)/(leading coefficient) = c(a-b)/[a(b-c)] = 1.</p><p><strong>Step 3:</strong> From Step 2: c(a-b) = a(b-c) → ca - cb = ab - ac → 2ac = ab + cb → 2ac = b(a+c).</p><p><strong>Step 4:</strong> Divide both sides by abc: 2ac/abc = b(a+c)/abc → 2/b = (a+c)/ac → 2/b = 1/c + 1/a.</p><p><strong>∴ Answer:</strong> 2/b = 1/a + 1/c</p>
Correct Answer: 2/b = 1/a + 1/c