Solve the radical equation. q minus 6 is equal to square root of 27 minus 2 q end root What is the extraneous solution to the radical equation?

Squaring both sides gives (q − 6)² = 27 − 2q, so q² − 10q + 9 = 0, which factors to (q − 1)(q − 9) = 0. Check both values in the original equation: q = 9 works, but q = 1 gives −5 = √25 = 5, which is false. Therefore, 1 is the extraneous solution.

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