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Advanced Math / Nonlinear equations in one variable and systems of equations in two variables Difficulty: Hard

$y = - 2.5$

$y = x^{2} + 8 x + k$

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 $k$ ?

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Explanation

The correct answer is $14$ . It's given by the first equation of the system of equations that $y = - 2.5$ . Substituting $negative 2.5$ 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, $b squared minus 4 a c$ , is negative. In the equation $0 = x^{2} + 8 x + k + 2.5$ , where $k$ is a positive integer constant, $a equals 1$ , $b equals 8$ , and $c = k + 2.5$ . Substituting $1$ for $a$ , $8$ for $b$ , and $k plus 2.5$ for $c$ in $b squared minus 4 a c$ 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 $4 left parenthesis k plus 2.5 right parenthesis$ to both sides of this inequality yields $64 less than 4 left parenthesis k plus 2.5 right parenthesis$ . Dividing both sides of this inequality by $4$ yields $16 less than k plus 2.5$ . Subtracting $2.5$ from both sides of this inequality yields $13.5 less than k$ . Since $k$ is a positive integer constant, the least possible value of $k$ is $14$ .