Question 130742
You don't actually say, but I'll presume that you need to solve this equation for x.


{{{(5-x)^(1/2)-1=x}}}


Remember that {{{a^(1/2)=sqrt(a)}}}, so your equation becomes:


{{{sqrt(5-x)-1=x}}}


Add 1 to both sides:


{{{sqrt(5-x)=x+1}}}


Square both sides:


{{{5-x=x^2+2x+1}}}


Put the equation in standard form {{{ax^2+bx+c=0}}}


{{{-x^2-2x-1+5-x=0}}}
{{{-x^2-3x+4=0}}}


Multiply by -1:


{{{x^2+3x-4=0}}}


Since {{{4*-1=-4}}} and {{{4-1=3}}}, this quadratic factors to:


{{{(x+4)(x-1)=0}}},


Therefore {{{x=-4}}} or {{{x=1}}}.  However, since this is a radical equation (meaning that the variable is under a radical sign), there is a possibility that by squaring both sides of the equation, we may have derived an equation that is not exactly equivalent to the original and have introduced an extraneous root.


Check the answers:


{{{(5-1)^(1/2)-1=1}}}
{{{(4)^(1/2)-1=1}}}
{{{2-1=1}}}:  True statement.  {{{x=1}}} is in the solution set.


{{{(5-(-4))^(1/2)-1=-4}}}
{{{(9)^(1/2)-1=-4}}}
{{{3-1=2<>-4}}}:  False statement.  {{{x=-4}}} is NOT in the solution set.


{{{x=1}}}, and that's my final answer, Regis.