Question 84·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

Type $c=0$ as a starting integer, then type both given equations. Desmos shows a line and a parabola meeting near $(0.69,5.34)$ and $(5.81,7.91)$; click each crossing to see its coordinates. Each intersection solves both equations, so this trial has two solutions. Increasing $c$ lifts the parabola, but the graph alone won’t give an exact cutoff.

At a shared point, both equations give the same $y$. Divide the line’s equation by $2$, then set its expression for $y$ equal to the parabola’s expression:

Expand the square to make a quadratic, an equation containing $x^2$:

Subtract $\frac{x}{2}+5$ from both sides:

To complete the square is to rewrite the $x$ terms as one squared expression. Move $c+4$ to the right:

Half of $-\frac{13}{2}$ is $-\frac{13}{4}$, and its square is $\frac{169}{16}$. Add that square to both sides:

Rewrite the left as one square:

Combine the constants on the right:

If the right side is positive, the square equation has two $x$-values, one from each sign of the square root. If it’s zero, there’s one; if it’s negative, there are none. At most one includes both one and zero solutions. So require:

Add $c$ to both sides:

The cutoff is $\frac{105}{16}=6+\frac{9}{16}$, so it lies between $6$ and $7$. The value $6$ still gives two solutions; because $c$ must be an integer, the least value at or above the cutoff is $7$. Grid in 7.