Question 130152
{{{sqrt(x + 63) + 9 = x}}} Start with the given equation



{{{sqrt(x + 63) = x-9}}} Subtract 9 from both sides



{{{x+63 = (x-9)^2}}} Square both sides



{{{x+63 = x^2-18x+81}}} Foil




{{{0=x^2-18x+81-x-63}}}  Subtract x from both sides.  Subtract 63 from both sides. 


{{{0=x^2-19x+18}}}  Combine like terms



{{{0=(x-18)(x-1)}}} Factor the right side (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)




Now set each factor equal to zero:

{{{x-18=0}}} or  {{{x-1=0}}} 


{{{x=18}}} or  {{{x=1}}}    Now solve for x in each case



So our possible solutions are 

 {{{x=18}}} or  {{{x=1}}} 



Now let's check our possible solutions



Let's check the first solution {{{x=18}}}


{{{sqrt(x + 63) + 9 = x}}} Start with the given equation}}} Start with the given equation



{{{sqrt(18 + 63) + 9 = 18}}}  Plug in {{{x=18}}}. 



{{{sqrt(81) + 9 = 18}}} Add



{{{9 + 9 = 18}}} Take the square root of 81 to get 9



{{{18 = 18}}} Add. Since the two sides of the equation are equal, this verifies the solution {{{x=18}}}.




--------------------------



Let's check the first solution {{{x=1}}}


{{{sqrt(x + 63) + 9 = x}}} Start with the given equation}}} Start with the given equation



{{{sqrt(1 + 63) + 9 = 1}}}  Plug in {{{x=1}}}. 



{{{sqrt(64) + 9 = 1}}} Add



{{{8 + 9 = 18}}} Take the square root of 64 to get 8



{{{17 = 18}}} Add. Since the two sides of the equation are <b>not</b> equal, this means that {{{x=1}}} is an extraneous solution





-----------------------------


Answer:



So our only solution is {{{x=18}}}