SAT Question 2c288148: Nonlinear Equations In One Variable And Systems Of Equations In Two Variables answer and explanation with Desmos solution

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 $58-x>0$. It follows that if a solution to the equation $k-x=(58-x{)}^2$ satisfies $58-x>0$, 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 $k-x=3,364-116x+x^2$. Adding $x$ to both sides of this equation yields $k=x^2-115x+3,364$. Subtracting $k$ from both sides of this equation yields $0=x^2-115x+(3,364-k)$. The number of solutions to a quadratic equation in the form $0=a{x}^2+bx+c$, where $a$, $b$, and $c$ are constants, can be determined by the value of the discriminant, $b^2-4ac$. Substituting $-115$ for $b$, $1$ for $a$, and $3,364-k$ for $c$ in $b^2-4ac$ yields $(-115{)}^2-4(1)(3,364-k)$, or $4k-231$. The equation $0=x^2-115x+(3,364-k)$ has exactly one real solution if the discriminant is equal to zero, or $4k-231=0$. Subtracting $231$ from both sides of this equation yields $4k=231$. Dividing both sides of this equation by $4$ yields $k=57.75$. Therefore, if $k=57.75$, then the equation $0=x^2-115x+(3,364-k)$ has exactly one real solution. Substituting $57.75$ for $k$ in this equation yields $0=x^2-115x+(3,364-57.75)$, or $0=x^2-115x+3,306.25$, 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 $57.5=x$. To check whether this solution satisfies $58-x>0$, the solution, $57.5$, can be substituted for $x$ in $58-x>0$, which yields $58-57.5>0$, or $0.5>0$. Since $0.5$ is greater than $0$, it follows that if $k=57.75$, or $4k=231$, then the given equation has exactly one real solution. If $4k<231$, then the discriminant, $4k-231$, is negative and the given equation has no solutions. Therefore, the minimum possible value of $4k$ is $231$.