Solve the radical equation q-6=√27-2q

q - 6 = sqrt (27 - 2q)

Squaring both sides:-

(q - 6) ^2 = 27 - 2q

q^2 - 12q + 36 = 27 - 2q

q^2 - 10q + 9 = 0

(q - 9) (q - 1) = 0

so there are 2 solutions q = 1 and q = 9.

Testing to see if the solutions are not extraneous:-

q = 9:-

9 - 6 = 3

sqrt (27-18) = 3 So q = 9 is a solution (answer)

q = 1

q - 6 = - 5

sqrt (27 - 2) = sqrt 25 = - 5 so q = 1 is also an answer