Question

how to solve: rootsqrx+3=x-3

Answer 1

I'm going to assume you need to solve,\[\sqrt{x+3}=x-3\]In that case, square both sides and collect terms to form the quadratic:\[x+3=(x-3)^2\]\[x+3=x^2-6x+9\]\[0=x^2-7x+6\]which factors as\[0=(x-1)(x-6)\]Hence, we have possible solutions, \[x=1,x=6\]We need to check each of these solutions with the original equation, since by squaring, we have introduced the possibility of spurious roots. For x=1, \[\sqrt{1+3}=2\]whereas on the right-hand side\[1-3=-2\]so x=1 is not a solution. For x=6,\[\sqrt{6+3}=3\] on the left-hand side, and \[6-3=3\]on the right-hand side. So x=6 is the solution.