Check the radical domain before reading or squaring the inequality

The square root requires 16−x²≥0, so −4≤x≤4. The right side is nonnegative; any solution of the strict comparison must also have x>0.

Sketch $y=x-\sqrt{16-x^2}$ on its real domain. Its endpoints are (−4,−4) and (4,4), with y-intercept −4. The source labels its zero x≈2.83 and reads 2.83<x≤4. This is a rounded graphical boundary; the exact zero is 2√2, giving $2\sqrt2<x\le4$.

Sketch y=x and $y=\sqrt{16-x^2}$, the upper semicircle of radius 4. The source labels height 4, endpoints (−4,0),(4,0), and the intersection input≈2.83. Select the line above the semicircle, retaining the allowed right endpoint: $2\sqrt2<x\le4$. The left endpoint is not a solution.

Because x>0 and the radical is nonnegative, squaring is equivalent here: $x^2>16-x^2$, so x²>8. Combine the positive-input condition with −4≤x≤4 to obtain $2\sqrt2<x\le4$. The strict crossing is excluded, but x=4 gives 4>0 and is included.