The question said solve and I have managed to get it this far, but I don't know what to do about the x at the end... what are the answers? 7 is obvious, but I have a random x still there too and I'm not sure what that means anymore...

√(2&5x+1) = x – 1
(√(2&5x+1))2 = (x – 1)2
5x + 1 = x2 – x – x + 1
5x + 1 = x2 – 2x + 1
-5x -5x
1 = x2 – 2x +1 – 5x
-1 -1
0 = x2 – 7x
0 = x(x – 7)

Question

The question said solve and I have managed to get it this far, but I don't know what to do about the x at the end... what are the answers?  7 is obvious, but I have a random x still there too and I'm not sure what that means anymore... 

√(2&5x+1) = x – 1
(√(2&5x+1))2 = (x – 1)2
5x + 1 = x2 – x – x + 1
5x + 1 = x2 – 2x + 1
-5x              -5x
1 = x2 – 2x +1 – 5x
-1               -1
0 = x2 – 7x 
0 = x(x – 7)

anonymous · Answer

zero product principle, if x(x-7)=0, then x=0 or x-7=0
so x=0, x=7

anonymous · Answer

Awesome! Thank you! It's been a couple years since I have to do even the simple stuff and I seem to have forgotten some of the basics! Thank you so much!! :)) That makes perfect sense!