Question 336199
{{{ sqrt( s+25 )  +5 = s}}} Start with the given equation.



{{{ sqrt( s+25 ) = s-5}}} Subtract 5 from both sides.



{{{s+25 = (s-5)^2}}} Square both sides.



{{{s+25 = s^2-10s+25}}} FOIL



{{{0 = s^2-10s+25-s-25}}} Get every term to one side.



{{{0=s^2-11s+0}}} Combine like terms.



Notice that the quadratic {{{s^2-11s+0}}} is in the form of {{{As^2+Bs+C}}} where {{{A=1}}}, {{{B=-11}}}, and {{{C=0}}}



Let's use the quadratic formula to solve for "s":



{{{s = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{s = (-(-11) +- sqrt( (-11)^2-4(1)(0) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-11}}}, and {{{C=0}}}



{{{s = (11 +- sqrt( (-11)^2-4(1)(0) ))/(2(1))}}} Negate {{{-11}}} to get {{{11}}}. 



{{{s = (11 +- sqrt( 121-4(1)(0) ))/(2(1))}}} Square {{{-11}}} to get {{{121}}}. 



{{{s = (11 +- sqrt( 121-0 ))/(2(1))}}} Multiply {{{4(1)(0)}}} to get {{{0}}}



{{{s = (11 +- sqrt( 121 ))/(2(1))}}} Subtract {{{0}}} from {{{121}}} to get {{{121}}}



{{{s = (11 +- sqrt( 121 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{s = (11 +- 11)/(2)}}} Take the square root of {{{121}}} to get {{{11}}}. 



{{{s = (11 + 11)/(2)}}} or {{{s = (11 - 11)/(2)}}} Break up the expression. 



{{{s = (22)/(2)}}} or {{{s =  (0)/(2)}}} Combine like terms. 



{{{s = 11}}} or {{{s = 0}}} Simplify. 



So the <u>possible</u> solutions are {{{s = 11}}} or {{{s = 0}}}



However, if you plug in {{{s=0}}} into the original equation, you'll get a contradiction. So {{{s=0}}} is NOT a solution.



So the only true solution is {{{s=11}}}



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Jim