HMMT 二月 2014 · 代数 · 第 8 题

HMMT February 2014 — Algebra — Problem 8

Find all real numbers k such that r + kr + r + 4 kr + 16 = 0 is true for exactly one real number r .

解析

Find all real numbers k such that r + kr + r + 4 kr + 16 = 0 is true for exactly one real number r . 9 9 9 9 9 1 Answer: ± (OR , − OR − , ) OR ± 2 OR ± 2 . 25 Any real quartic has an even number of 4 4 4 4 4 4 4 3 2 real roots with multiplicity , so there exists real r such that x + kx + x + 4 kx + 16 either takes the 4 2 2 2 form ( x + r ) (clearly impossible) or ( x + r ) ( x + ax + b ) for some real a, b with a < 4 b . Clearly 2 16 32 2 2 r − 4 8 r 6 = 0, so b = and 4 k = 4( k ) yields + ar = 4(2 r + a ) = ⇒ a ( r − 4) = 8 . Yet a 6 = (or else 2 r r r r 2 2 2 16 − 7 7 9 a = 4 b ), so r = 4, and 1 = r + 2 ra + = ⇒ a = . Thus k = 2 r − = ± (since r = ± 2). 2 r 2 r 2 r 4 9 4 3 2 1 2 2 It is easy to check that k = works, since x + (9 / 4) x + x + 4(9 / 4) x + 16 = ( x + 2) (4 x − 7 x + 16). 4 4 9 1 2 2 The polynomial given by k = − is just ( − x + 2) (4 x + 7 x + 16). 4 4 2 4 3 2 2 k 2 k 2 Alternate solution: x + kx + x + 4 kx + 16 = ( x + x + 4) + (1 − 8 − ) x , so for some ǫ ∈ {− 1 , 1 } , 2 4 √ 2 2 2 2 x +( k − ǫ k + 28) x +8 has a single real root and thus takes the form 2( x + r ) (using the same notation √ √ 2 2 2 2 2 2 as above). But then ( k − ǫ k + 28) = 4(2)(8) = 8 , so we conclude that ( k ± 8) = ( ǫ k + 28) 7 9 and k = ± (4 − ) = ± . 4 4