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

$\sqrt{k - x} = 58 - x$

In the given equation, $k$ is a constant. The equation has exactly one real solution. What is the minimum possible value of $4 k$ ?

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Explanation

The correct answer is $231$ . It's given that $\sqrt{k - x} = 58 - x$ . Squaring both sides of this equation yields $k - x = (58 - x)^{2}$ , which is equivalent to the given equation if $58 minus x greater than 0$ . It follows that if a solution to the equation $k - x = (58 - x)^{2}$ satisfies $58 minus x greater than 0$ , then it's also a solution to the given equation; if not, it's extraneous. The equation $k - x = (58 - x)^{2}$ can be rewritten as $k - x = 3,364 - 116 x + x^{2}$ . Adding $x$ to both sides of this equation yields $k = x^{2} - 115 x + 3,364$ . Subtracting $k$ from both sides of this equation yields $0 = x^{2} - 115 x + (3,364 - k)$ . The number of solutions to a quadratic equation in the form $0 = a x^{2} + b x + c$ , where $a$ , $b$ , and $c$ are constants, can be determined by the value of the discriminant, $b squared minus 4 a c$ . Substituting $negative 115$ for $b$ , $1$ for $a$ , and $3,364 minus k$ for $c$ in $b squared minus 4 a c$ yields $(- 115)^{2} - 4 (1) (3,364 - k)$ , or $4 k minus 231$ . The equation $0 = x^{2} - 115 x + (3,364 - k)$ has exactly one real solution if the discriminant is equal to zero, or $4 k - 231 = 0$ . Subtracting $231$ from both sides of this equation yields $4 k equals 231$ . Dividing both sides of this equation by $4$ yields $k equals 57.75$ . Therefore, if $k equals 57.75$ , then the equation $0 = x^{2} - 115 x + (3,364 - k)$ has exactly one real solution. Substituting $57.75$ for $k$ in this equation yields $0 = x^{2} - 115 x + (3,364 - 57.75)$ , or $0 = x^{2} - 115 x + 3,306.25$ , which is equivalent to $0 = (x - 57.5)^{2}$ . Taking the square root of both sides of this equation yields $0 = x - 57.5$ . Adding $57.5$ to both sides of this equation yields $57.5 equals x$ . To check whether this solution satisfies $58 minus x greater than 0$ , the solution, $57.5$ , can be substituted for $x$ in $58 minus x greater than 0$ , which yields $58 minus 57.5 greater than 0$ , or $0.5 greater than 0$ . Since $0.5$ is greater than $0$ , it follows that if $k equals 57.75$ , or $4 k equals 231$ , then the given equation has exactly one real solution. If $4 k less than 231$ , then the discriminant, $4 k minus 231$ , is negative and the given equation has no solutions. Therefore, the minimum possible value of $4 k$ is $231$ .