It depends if you mean \(\sqrt{8x} +1=5\) or \(\sqrt{8x+1} = 5\)
\(\sqrt{8x} +1=5\)
Subtract 1 from both sides.
\(\sqrt{8x} =4\)
Take the square root of both sides.
8x = sqrt(4)
8x = 2
x = 1/4.
Plugging it back in:
sqrt(8*0.25) + 1 = 5
sqrt(2) = 4
This is not true because the square root of 2 is about 1.41, which does not equal 4.
So it is an extraneous solution.
__________________________________________
\(\sqrt{8x+1} = 5\)
Square both sides.
8x + 1 = 25
8x = 24
x = 3
Not extraneous because equation works out when you plug x back in.
