## College Algebra (6th Edition)

$x=8$
Square both sides and put all terms on one side. Divide both sides by $x$. One of the answers does not work. $\sqrt {6x+1}=x-1$ $(\sqrt {6x+11})^2=(x-1)^2$ $6x+1=x^2-2x+1$ $x^2-8x=0$ $x^2=8x$ $x=0,8$ $\sqrt{6\times0+1}\ne0-1$ $x=8$