\( 1 \leftarrow \quad \) Solve. \( \quad \sqrt{32-x}=x-2 \)

To solve the equation \( \sqrt{32 - x} = x - 2 \), start by squaring both sides to eliminate the square root. This gives you \( 32 - x = (x - 2)^2 \). Expanding the right side results in \( 32 - x = x^2 - 4x + 4 \). Rearranging all terms leads to the standard quadratic equation: \( x^2 - 3x - 28 = 0 \). You can factor this equation as \( (x - 7)(x + 4) = 0 \). Thus, the solutions are \( x = 7 \) and \( x = -4 \). However, don't forget to check for extraneous solutions by plugging both values back into the original equation. Checking \( x = 7 \): \( \sqrt{32 - 7} = 5 \) and \( 7 - 2 = 5 \), so it's valid! Checking \( x = -4 \): \( \sqrt{32 - (-4)} = 6\) and \( -4 - 2 = -6\) (not valid). Hence, the only solution is \( x = 7 \).