Question 96890
{{{x+3=sqrt(x+15)}}} Start with the given equation



{{{(x+3)^2=x+15}}} Square both sides



{{{x^2+6x+9=x+15}}} Foil



{{{x^2+5x-6=0}}} Get all terms to one side





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



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}


So lets solve {{{x^2+5*x-6=0}}} ( notice {{{a=1}}}, {{{b=5}}}, and {{{c=-6}}})


{{{x = (-5 +- sqrt( (5)^2-4*1*-6 ))/(2*1)}}} Plug in a=1, b=5, and c=-6




{{{x = (-5 +- sqrt( 25-4*1*-6 ))/(2*1)}}} Square 5 to get 25  




{{{x = (-5 +- sqrt( 25+24 ))/(2*1)}}} Multiply {{{-4*-6*1}}} to get {{{24}}}




{{{x = (-5 +- sqrt( 49 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-5 +- 7)/(2*1)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{x = (-5 +- 7)/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (-5 + 7)/2}}} or {{{x = (-5 - 7)/2}}}


Lets look at the first part:


{{{x=(-5 + 7)/2}}}


{{{x=2/2}}} Add the terms in the numerator

{{{x=1}}} Divide


So one answer is

{{{x=1}}}




Now lets look at the second part:


{{{x=(-5 - 7)/2}}}


{{{x=-12/2}}} Subtract the terms in the numerator

{{{x=-6}}} Divide


So another answer is

{{{x=-6}}}


So our possible solutions are:

{{{x=1}}} or {{{x=-6}}}




Let's check them:



Let's check the solution x=-6

{{{x+3=sqrt(x+15)}}} Start with the given equation



{{{-6+3=sqrt(-6+15)}}} Plug in x=-6



{{{-3=sqrt(9)}}} Combine like terms



{{{-3=3}}} Take the square root of 9. Since this is not true, x=-6 is not a solution (it's extraneous)



Let's check the solution x=1


{{{x+3=sqrt(x+15)}}} Start with the given equation



{{{1+3=sqrt(1+15)}}} Plug in x=1



{{{4=sqrt(16)}}} Combine like terms



{{{4=4}}} Take the square root of 16. Since this is true, x=1 is a solution 




So our only solution is {{{x=1}}}