y = - 2.5 y = x 2 + 8 x + k In the given system of equations, k

Question

y = - 2.5 y = x 2 + 8 x + k 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 ?

Accepted Answer

The correct answer is 14 . It's given by the first equation of the system of equations that y = - 2.5 . Substituting - 2.5 for y in the second given equation, y = x 2 + 8 x + k , yields - 2.5 = x 2 + 8 x + k . Adding 2.5 to both sides of this equation yields 0 = x 2 + 8 x + k + 2.5 . A quadratic equation of the form 0 = a x 2 + b x + c , where a , b , and c are constants, has no real solutions if and only if its discriminant, b 2 - 4 a c , is negative. In the equation 0 = x 2 + 8 x + k + 2.5 , where k is a positive integer constant, a = 1 , b = 8 , and c = k + 2.5 . Substituting 1 for a , 8 for b , and k + 2.5 for c in b 2 - 4 a c yields 8 2 - 4 1 k + 2.5 , or 64 - 4 k + 2.5 . Since this value must be negative, 64 - 4 k + 2.5 < 0 . Adding 4 k + 2.5 to both sides of this inequality yields 64 < 4 k + 2.5 . Dividing both sides of this inequality by 4 yields 16 < k + 2.5 . Subtracting 2.5 from both sides of this inequality yields 13.5 < k . Since k is a positive integer constant, the least possible value of k is 14 .

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