Question 87636
{{{(2x-1)^(1/2)=x-2}}}


Square both sides:

{{{2x-1= (x-2)^2}}}
{{{2x-1 = x^2 -4x+4}}}


It's a quadratic equation, so set equal to zero by subtracting 2x and adding +1 to each side of the equation:
{{{2x-1 -2x+1 = x^2 -4x +4 -2x +1 }}}
{{{0=x^2 -6x +5}}}


Which FACTORS!!  (They all do when solving radical equations!!)
{{{0=(x-5)(x-1) }}}
{{{x=5}}} or {{{x=1}}}


Answers are NOT guaranteed, so you need to check the answers in the original equation to see if they work!


x=5
{{{(2x-1)^(1/2)=x-2}}}
{{{(10-1)^(1/2)=5-2}}}  It checks!!


x=1
{{{(2x-1)^(1/2)=1-2}}}
{{{(2-1)^(1/2)=x-2}}}
{{{1^(1/2) = -1}}}  Reject.  This does NOT check!!


The answer is x=5.


R^2 Retired from SCC