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

$x^{2} - 34 x + c = 0$

In the given equation, $c$ is a constant. The equation has no real solutions if $c > n$ . What is the least possible value of $n$ ?

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Explanation

The correct answer is $289$ . A quadratic equation of the form $a x^{2} + b x + c = 0$ , where $a$ , $b$ , and $c$ are constants, has no real solutions when the value of the discriminant, $b squared minus 4 a c$ , is less than $0$ . In the given equation, $x^{2} - 34 x + c = 0$ , $a equals 1$ and $b = - 34$ . Therefore, the discriminant of the given equation can be expressed as ${(- 34)}^{2} - 4 (1) (c)$ , or $1,156 minus 4 c$ . It follows that the given equation has no real solutions when $1,156 minus 4 c less than 0$ . Adding $4 c$ to both sides of this inequality yields $1,156 less than 4 c$ . Dividing both sides of this inequality by $4$ yields $289 less than c$ , or $c greater than 289$ . It's given that the equation $x^{2} - 34 x + c = 0$ has no real solutions when $c > n$ . Therefore, the least possible value of $n$ is $289$ .