For x2 − Sx + P = 0 with nonzero roots r1, r2, find Σ 1/ri and Σ 1/ri2.
Solution
Correct Answer: Option C
Correct answer: Option 1. Explanation (textbook style).
Let the quadratic be x^2 − Sx + P = 0 with nonzero roots r1, r2. By Viète’s relations
r1 + r2 = S, r1 r2 = P.
Hence the sum of reciprocals is
Σ 1/r_i = 1/r1 + 1/r2 = (r1 + r2)/(r1 r2) = S/P.
To get the sum of reciprocal squares,
Σ 1/r_i^2 = 1/r1^2 + 1/r2^2 = (r1^2 + r2^2)/(r1^2 r2^2).
But r1^2 + r2^2 = (r1 + r2)^2 − 2r1 r2 = S^2 − 2P, and r1^2 r2^2 = (r1 r2)^2 = P^2. Therefore
Σ 1/r_i^2 = (S^2 − 2P)/P^2.
(Equivalently, letting y = 1/x and dividing the original equation by x^2 gives P y^2 − S y + 1 = 0, so the sum of the roots y = S/P and the sum of their squares equals (S^2 − 2P)/P^2.)
Example (to check): x^2 − 2x − 1 = 0 has S = 2, P = −1. Then Σ 1/r_i = 2/(−1) = −2 and Σ 1/r_i^2 = (4 − 2(−1))/1 = 6 (which matches a direct computation of the reciprocals).
Let the quadratic be x^2 − Sx + P = 0 with nonzero roots r1, r2. By Viète’s relations
r1 + r2 = S, r1 r2 = P.
Hence the sum of reciprocals is
Σ 1/r_i = 1/r1 + 1/r2 = (r1 + r2)/(r1 r2) = S/P.
To get the sum of reciprocal squares,
Σ 1/r_i^2 = 1/r1^2 + 1/r2^2 = (r1^2 + r2^2)/(r1^2 r2^2).
But r1^2 + r2^2 = (r1 + r2)^2 − 2r1 r2 = S^2 − 2P, and r1^2 r2^2 = (r1 r2)^2 = P^2. Therefore
Σ 1/r_i^2 = (S^2 − 2P)/P^2.
(Equivalently, letting y = 1/x and dividing the original equation by x^2 gives P y^2 − S y + 1 = 0, so the sum of the roots y = S/P and the sum of their squares equals (S^2 − 2P)/P^2.)
Example (to check): x^2 − 2x − 1 = 0 has S = 2, P = −1. Then Σ 1/r_i = 2/(−1) = −2 and Σ 1/r_i^2 = (4 − 2(−1))/1 = 6 (which matches a direct computation of the reciprocals).
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