Let a, b, c be the lengths of three sides of a triangle satistying the condition (a²+b²) x²-2 b(a+c) x+(b²+c²)=0. If the set of all possible values of x is the interval (α, β), then 12(α²+β²) is equal to __________.

Solution

Maths sir · 4 steps

Split the quadratic into two perfect squares; then the sides are in GP, and the triangle inequality bounds x.

1. Regroup the terms. Expand and pair the terms: a²x² - 2abx + b² and b²x² - 2bcx + c². Each pair is a perfect square.

   (ax - b)² + (bx - c)² = 0

2. Both squares vanish. For real x, a sum of two squares is zero only if each square is zero. So x = b/a = c/b, which gives b = ax and c = ax². The sides a, ax, ax² are in GP with ratio x > 0.
3. Triangle inequality. Divide each inequality by a. From 1 + x > x² we get x² - x - 1 < 0, so x < (1+√5)/2. From x + x² > 1 we get x > (√5-1)/2. The third inequality, 1 + x² > x, holds for every real x.
4. So. α = (√5-1)/2 and β = (√5+1)/2. Then α² + β² = ((6 - 2√5) + (6 + 2√5))/4 = 3, so 12(α² + β²) = 36.

The idea to keep. When a quadratic has symmetric-looking coefficients built from squares, try writing it as a sum of squares equal to zero. Each square must then vanish, and that gives you the relation directly.

Answer from the NTA final answer key. Solution by PrepWiser.