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

It depends if you mean  or  



Subtract 1 from both sides. 



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.
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Square both sides. 
8x + 1 = 25 
8x = 24
x = 3

Not extraneous because equation works out when you plug x back in.

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