Let α and β be the roots of the equation p x²+q x-r=0, where p ≠ 0. If p, q and r be the consecutive terms of a non constant G.P. and 1/α+1/β=3/4, then the value of (α-β)² is :

Solution

Maths sir · 4 steps

Write everything in terms of the common ratio; the given condition fixes it, and the rest follows.

1. Sum and product. For px² + qx - r = 0, α + β = -q/p and αβ = -r/p. Watch the minus sign on the constant term.
2. Use the condition. 1/α + 1/β = (α + β)/αβ = (-q/p)/(-r/p) = q/r. So q/r = 3/4.
3. Find the ratio. Let the common ratio be k, so q = pk and r = pk². Then q/r = 1/k = 3/4, which gives k = 4/3.

   (α-β)² = (α+β)² - 4αβ = q²/p² + 4r/p = k² + 4k² = 5k²

4. So. (α - β)² = 5 × 16/9 = 80/9, which is option (d).

The idea to keep. When coefficients are in a G.P., put them as p, pk, pk². Every root expression then depends on k alone, and (α-β)² = (α+β)² - 4αβ finishes the job.

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