Solve $\sqrt {x - 15} = 3 - \sqrt x $ and check the solution?

Hint: First take square on both the sides, then we will rearrange the terms so that the terms with x are cancelled. Then, we will get an equation in x which can be solved for x.

Complete step by step answer:
We are given that we are required to solve $\sqrt {x - 15} = 3 - \sqrt x $ and check the solution.
Let us take square of both the sides of the equation above, we will then obtain the following expression
$ \Rightarrow {\left( {\sqrt {x - 15} } \right)^2} = {\left( {3 - \sqrt x } \right)^2}$
Simplifying the left hand side of the above equation, we will then obtain the following equation
$ \Rightarrow x - 15 = {\left( {3 - \sqrt x } \right)^2}$ ………………(1)
Now, we also know that we have an identity given by the following expression
$ \Rightarrow {\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
Replacing a by 3 and b by $\sqrt x $ in the above mentioned formula, we will then obtain the following equation
$ \Rightarrow {\left( {3 - \sqrt x } \right)^2} = {3^2} + {\left( {\sqrt x } \right)^2} - 2\left( 3 \right)\left( {\sqrt x } \right)$
Simplifying the calculations on the right hand side, we will then obtain the following equation
$ \Rightarrow {\left( {3 - \sqrt x } \right)^2} = 9 + x - 6\sqrt x $
Putting this in equation number 1, we will then obtain the following equation
$ \Rightarrow x - 15 = 9 + x - 6\sqrt x $
Taking x from addition in the right hand side to subtraction in the left hand side, we will then obtain the following equation
$ \Rightarrow x - 15 - x = 9 - 6\sqrt x $
Simplifying the calculations on the left hand side, we will then obtain the following equation
$ \Rightarrow - 15 = 9 - 6\sqrt x $
Taking 9 from addition in the right hand side to subtraction in the left hand side, we will then obtain the following equation
$ \Rightarrow - 15 - 9 = - 6\sqrt x $
Simplifying the calculations on the left hand side, we will then obtain the following equation
$ \Rightarrow - 24 = - 6\sqrt x $
Multiplying both the sides of above equation by -1 and dividing by 6 on both the sides, we will get the following expression
$ \Rightarrow \sqrt x = 4$
Taking square on both the sides, we will then obtain
$ \Rightarrow x = 16$
Now, to check it let us put x = 16 in $\sqrt {x - 15} = 3 - \sqrt x $.
Left hand side: $\sqrt {16 - 15} = \sqrt 1 = \pm 1$
Right hand side: $3 - \sqrt {16} = 3 - 4 = - 1$

Note: The students must note that it could have been the case that the x in the equation was not cancelled due to difference in coefficient, then we would have approached it in the following way
We will replace ${y^2} = x$ so that we have $y = \sqrt x $ and thus we have a quadratic equation in y which can be easily solved using the quadratic formula or the splitting of the middle term to get the value of y and thus the value for x.
The students must commit to the memory the following formulas
$ \Rightarrow {\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$