Given the equation Square root of 8x plus 1 = 5, solve for x and identify if it is an extraneous solution.

do recall that squaring and the *radical sign* cancel each other out... like so:([tex] \sqrt{a} [/tex])[tex] ^{2} [/tex]= a

When you put it that way, it isn't enough :P
([tex] \sqrt{a} [/tex])[tex] ^{2} [/tex]= a
([tex] \sqrt{8x+1} [/tex])[tex] ^{2} [/tex]=?

so you start with
([tex] \sqrt{8x+1} [/tex])[tex] ^{2} [/tex]= [tex] (5)^{2} [/tex]
8x+1=25 <-- subtract 1 to both sides
8x=24 <- divide 8 to both sides
x= 3

To find out if it's an extraneous solution ask yourself: It mustn't result in a radical that I like to call... 'illegal'. Plug it into the radicand 8x+1 and make sure you get something that is not a negative number.... so, DO you get a negative number when you plug in x = 3 into the radicand?

(extraneous solution is a invalid solution)

x=3 not extraneous