Assuming the equation is a genuine quadratic with non-zero roots, find the sum…
2019
Assuming the equation is a genuine quadratic with non-zero roots, find the sum of the reciprocals of the roots of abx2 = (a2 + b2 + 2ab)(x - 1).
- A.
2/3
- B.
1
- C.
2
- D.
1/2
Attempted by 4 students.
Show answer & explanation
Correct answer: B
Concept
For a quadratic Ax2 + Bx + C = 0 with non-zero roots α and β, Vieta's formulas give α + β = -B/A and αβ = C/A.
Therefore, the sum of the reciprocals is 1/α + 1/β = (α + β)/(αβ) = -B/C. This requires A and C to be non-zero.
Application
Expand the bracket: a2 + b2 + 2ab = (a + b)2.
Move every term to one side: abx2 - (a + b)2x + (a + b)2 = 0.
Here A = ab, B = -(a + b)2, and C = (a + b)2.
Apply -B/C: -[-(a + b)2]/(a + b)2 = 1.
Cross-check
Vieta also gives α + β = (a + b)2/(ab) and αβ = (a + b)2/(ab). Their ratio is 1, confirming the result.
Hence, the sum of the reciprocals of the roots is 1.