It depends if you mean [tex] \sqrt{8x} +1=5[/tex] or [tex] \sqrt{8x+1} = 5 [/tex]
[tex] \sqrt{8x} +1=5[/tex]
Subtract 1 from both sides.
[tex] \sqrt{8x} =4[/tex]
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. __________________________________________ [tex] \sqrt{8x+1} = 5 [/tex] Square both sides. 8x + 1 = 25 8x = 24 x = 3
Not extraneous because equation works out when you plug x back in.
