Answer:
r = 3, solution is not extraneous
Explanation:
I assume you mean
√(8r + 1) = 5
Also, why is the variable in the equation r,
and all solution choices have x = ...
Solving the equation:
√(8r + 1) = 5
Square both sides.
8r + 1 = 5²
8r + 1 = 25
Subtract 1 from both sides.
8r = 24
Divide both sides by 8.
r = 3
Check r = 3 in the original equation.
√(8r + 1) = 5
√(8 × 3 + 1) = 5
√(24 + 1) = 5
√(25) = 5
5 = 5
r = 3 works in the original equation, so the solution is r = 3.
r = 3 is not extraneous.