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

$x (k x - 56) = - 16$

In the given equation, $k$ is an integer constant. If the equation has no real solution, what is the least possible value of $k$ ?

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Explanation

The correct answer is $50$ . An equation of the form $a x^{2} + b x + c = 0$ , 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. Applying the distributive property to the left-hand side of the equation $x (k x - 56) = - 16$ yields $k x^{2} - 56 x = - 16$ . Adding $16$ to each side of this equation yields $k x^{2} - 56 x + 16 = 0$ . Substituting $k$ for $a$ , $negative 56$ for $b$ , and $16$ for $c$ in $b squared minus 4 a c$ yields a discriminant of ${(- 56)}^{2} - 4 (k) (16)$ , or $3,136 minus 64 k$ . If the given equation has no real solution, it follows that the value of $3,136 minus 64 k$ must be negative. Therefore, $3,136 minus 64 k less than 0$ . Adding $64 k$ to both sides of this inequality yields $3,136 less than 64 k$ . Dividing both sides of this inequality by $64$ yields $49 less than k$ , or $k greater than 49$ . Since it's given that $k$ is an integer, the least possible value of $k$ is $50$ .