Let a, b be two non-zero real numbers. If p and r are the roots of the equation x²-8 a x+2 a=0 and q and s are the roots of the equation x²+12 b x+6 b=0, such that 1/p, 1/q, 1/r, 1/s are in A.P., then a⁻¹-b⁻¹ is equal to _____________.

Solution

Maths sir · 4 steps

Reciprocals of roots? Flip each equation with x = 1/y and read off the new sum and product.

1. Reciprocal equations. Put x = 1/y in x² - 8ax + 2a = 0 to get 2ay² - 8ay + 1 = 0, whose roots are 1/p, 1/r. Their sum is 4 and their product is 1/2a. The same move on x² + 12bx + 6b = 0 gives 6by² + 12by + 1 = 0, so 1/q + 1/s = -2 and 1/q · 1/s = 1/6b.
2. Write the A.P. Let the terms be α, α + d, α + 2d, α + 3d. The first and third are 1/p, 1/r, so 2α + 2d = 4, which gives α + d = 2. The second and fourth are 1/q, 1/s, so 2α + 4d = -2, which gives α + 2d = -1.
3. Find the terms. Subtracting the first equation from the second gives d = -3, and then α = 5. So 1/p, 1/q, 1/r, 1/s are 5, 2, -1, -4.

   1/2a = 5 × (-1) = -5 ⇒ a⁻¹ = -10, 1/6b = 2 × (-4) = -8 ⇒ b⁻¹ = -48

4. So. a⁻¹ - b⁻¹ = -10 - (-48) = 38.

The common slip: -38. Subtracting in the wrong order, b⁻¹ - a⁻¹, gives -38, not 38.

The idea to keep. When a question talks about reciprocals of roots, flip the equation first with x = 1/y. The new sum and product are then ready to plug into the A.P. conditions.

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