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= −2.5 y = +8x+k y x^2 In the given system of equations, k is a positive integer constant. The system has no real solutions. What is the least possible value of k? OP
Correct answer: 14 (parsed from the explanation — this source omits an answer key for this item)
Explanation
14 The correct answer is . It’s given by the first equation of the system of =-2.5 -2.5 equations that y . Substituting for y in the second given equation, = 2+8 + -2.5= 2+8 + 2.5 y x x k, yields x x k. Adding to both sides of this 0= 2+8 + +2.5 equation yields x x k . A quadratic equation of the form 0= 2+ + ax bx c, where a, b, and c are constants, has no real solutions if and only 2-4 0= 2+8 + +2.5 if its discriminant, b ac, is negative. In the equation x x k , = = = +2.5 where k is a positive integer constant, a 1, b 8, and c k . Substituting 1 +2.5 2-4 82-4 1 +2.5 for a, 8 for b, and k for c in b ac yields ^ h^k h, or 64-4 +2.5 64-4 +2.5 <0 ^k h. Since this value must be negative, ^k h . Adding 4 +2.5 64<4 +2.5 ^k h to both sides of this inequality yields ^k h. Dividing both 16< +2.5 2.5 sides of this inequality by 4 yields k . Subtracting from both sides of 13.5< this inequality yields k. Since k is a positive integer constant, the least 14 possible value of k is . 46 SAT PRACTICE TEST #7 ANSWER EXPLANATIONS
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