<p>If \(y = \sqrt{28+\sqrt{28+\sqrt{28+\cdots}}}\) (where \(y > 0\)), then which of the following is/are correct?</p>
Step-by-Step Solution
Key Concept: Let y = \sqrt{28+y}. Square: y^2 = 28+y \to y^2-y-28 = 0 \to (y-7)(y+4) = 0 \to y = 7 (positive). Verify each option.
Notice that the best first move is to reveal the hidden structure in the expression. A clever move here is to rewrite the problem in the form where the standard theorem or identity applies cleanly. \(y=\sqrt{28+y}\) (A ✓). Squaring: \(y^2=28+y\Rightarrow y^2-y-28=0\) (C ✓). Factor: \((y-7)(y+4)=0\Rightarrow y=7\) (not 14, so B ✗). Also \(y^2-y=28\Rightarrow y(y-1)=28\) (D ✓). Answers: A, C, D. Now, we invoke the power of that idea, simplify patiently, and then check that the final answer really fits the original problem.
Correct Answer: A, C, D