(1) When x=-1, we get √6, and x+1=0, and √6>0.
When x=5, we get 0 (LHS) and x+1=6, and 0<6.
Note that we have a change of sign implying that 0 comes somewhere in between. That means there must be a value of x (=c) between -1 and 5 such that √(5-c)=c+1.
(2) Square both sides:
There appear to be two roots, x=1 and x=-4.
(3) We know from (1) that the solution is between -1 and 5, so x=1 must be the solution, because it lies between these two values.
(Why isn’t x=-4 a solution? Plug x=-4 into the original equation and we get:
√9=-3. Certainly, (-3)²=9, but the question specifies an implied positive square root and -3 is negative, so we throw out x=-4 as a solution.)