Question 35509
{{{sqrt(x+7) = x+1) }}}


Square both sides, which eliminates the radical on the left side:
{{{x+7 = (x+1)^2}}}
{{{x+7 = x^2 + 2x + 1}}}


This is quadratic, so set it equal to zero by subtracting x and 7 from each side.
{{{0 = x^2 +2x +1 - x - 7}}}
{{{0 = x^2 + x - 6}}}


It factors!!!  (I think they ALWAYS do!!)
{{{ 0= (x+3)(x-2) }}}
{{{x = -3}}}   {{{  x= 2}}}


HOWEVER, you must check these answers, since you squared both sides of the equation.  Substitute both values back into the original equation.


Check x = -3:
{{{sqrt(x+7) = x+1) }}}
{{{sqrt(-3+7) = -3+1}}}
{{{sqrt(4) = -2}}}  
This answer must be rejected by the principle value definition of square root.


Check x = 2:
{{{sqrt(x+7) = x+1) }}}
{{{sqrt(2+7) = 2+1}}}
{{{sqrt(9) = 3}}}  
This answer checks!!


Final answer is x= 2.


R^2 at SCC