After you set $x + 1 = \sqrt{2x + k}$, square both sides to remove the square root, but remember that a square root is always nonnegative. What condition does that put on $x + 1$?

Your squared equation should simplify to something like $x^2 + 1 = k$. For a given $k$, how many real $x$ values satisfy this, and which of them also meet the condition you found for $x$?

Use your relationship between $x$ and $k$ to quickly check each option. For each $k$, decide whether there are zero, one, or two valid $x$ values, and look for the option that gives exactly one.

In Desmos, enter y = sqrt(2x + k) on one line and y = x + 1 on another. Desmos will prompt you to create a slider for k; create that slider.

Move the k slider to each of the values in the choices ($k = -3, 0, 2, 5$). For each setting, look at the graph and count how many intersection points appear between the line and the square-root curve.

Among the tested values of k, note which one makes the graphs intersect at exactly one point. That value of k is the correct answer to the problem.

From the system

any solution must satisfy both equations at once. That means

Because a square root is always nonnegative, $\sqrt{2x + k} \ge 0$, so we must also have $x + 1 \ge 0$, or

This inequality will be important when we count how many real solutions are actually valid.

To remove the square root, square both sides of

This gives

Expand and simplify:

So any $x$-coordinate of a solution must satisfy

together with $x \ge -1$ from Step 1.

From $x^2 + 1 = k$ we have

Now consider cases:

• If $k < 1$, then $k - 1 < 0$ and $x^2 = k - 1$ has no real solutions.
• If $k = 1$, then $x^2 = 0$ so $x = 0$, which satisfies $x \ge -1$, giving one solution.
• If $k > 1$, then $x^2 = k - 1$ has two real solutions

Now apply the condition $x \ge -1$:

• The positive root $x = +\sqrt{k - 1}$ is always $\ge 0$, so it is always allowed.
• For the negative root $x = -\sqrt{k - 1}$ to be allowed, we need

So:

• If $1 < k \le 2$, both roots are valid (two solutions).
• If $k > 2$, only the positive root is valid (one solution).

Use the summary from Step 3:

• $k = -3$ and $k = 0$ are both less than 1, so there are no real solutions.
• $k = 2$ is in the range $1 < k \le 2$, so there are two real solutions.
• $k = 5$ is greater than 2, so there is exactly one valid real solution.

Therefore, the value of $k$ for which the system has exactly one real solution is $\boxed{5}$.