Question 165·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

Type $k=-3$, $y=\sqrt{2x+k}$, and $y=x+1$ on separate Desmos lines. An intersection is a point on both graphs, so each one is a solution. Desmos shows no intersections at this setting. Count only points shared by the original graphs.

Change the $k$ line to $k=0$. Desmos still shows no shared point, so this setting gives no real solution.

Change the $k$ line to $k=2$ and click both intersections. Desmos labels them $(-1,0)$ and $(1,2)$, so there are two solutions. Don't miss $(-1,0)$: the square root equals $0$ there, and the point is also on the line.

Change the $k$ line to $k=5$. The graphs share one point; click it to read $(2,3)$. So $k=5$ gives exactly one real solution to the system. Choice D.

No Desmos needed. The square-root sign tells us which candidates can work. A square root is never negative. Since it equals $x+1$, any solution must satisfy:

Subtract $1$:

At a solution, both equations give the same $y$, so:

Square both sides to get an equation without the radical:

Squaring can add candidates, so we'll still use the restriction $x\ge-1$.

Expand the square:

Subtract $2x$:

Subtract $1$:

Now the possible $x$-values depend on whether $k-1$ is negative, zero, or positive.

A real number's square cannot be negative. At $k=-3$ and $k=0$, the equation becomes $x^2=-4$ and $x^2=-1$, so neither works. At $k=2$, it becomes $x^2=1$, giving $x=-1$ and $x=1$. Both pass $x\ge-1$, so that setting gives two solutions.

At $k=5$, the equation is $x^2=4$, giving $x=-2$ or $x=2$. But $-2$ fails $x\ge-1$: it makes $x+1$ negative, which can't equal a square root. Only $x=2$ works, and the line gives $y=3$. So this setting gives exactly one real solution, $(2,3)$. Choice D.