How do you solve $ \sqrt x - \sqrt {x - 5} = 1 $ ?

Hint: In order to solve and write the expression into the simplest form . The square root is related to figuring out what should be the number which when multiplied by itself is equal to the number under the square root symbol $ \sqrt {} $ . This symbol is known as radical . Since in our case we have given the question in which we have to solve and find the value of x , we will first get rid of the radical and remove the square root as we want the original value of x , by somewhere using equivalent equations . Equivalent equations are said to be algebraic equations that may have the same solutions if we add or subtract the same number to both sides of an equation - Left hand side or Right hand side of the equal to sign . Or we can multiply or divide the same number to both sides of an equation - Left hand side or Right hand side of the equal to sign .

Complete step-by-step answer:
If we see the question , we need to solve the given expression under the square root which is $ \sqrt x - \sqrt {x - 5} = 1 $ . -----equation 1
By applying the concept of equivalent equation , we will first multiply both sides with $ \sqrt x + \sqrt {x - 5} = 1 $ both the L . H . S . and the R . H . S . as follows –
We are going to multiply the both by $ \sqrt x + \sqrt {x - 5} = 1 $ . So that we can apply the identity as stated below –
 $ \left( {\sqrt x - \sqrt {x - 5} } \right)\left( {\sqrt x + \sqrt {x - 5} } \right) = 1\left( {\sqrt x + \sqrt {x - 5} } \right) $
We can apply the formula or we can say identity $ (a + b)(a - b) = {a^2} - {b^2} $
 $
  {\left( {\sqrt x } \right)^2} - {\left( {\sqrt {x - 5} } \right)^2} = \sqrt x + \sqrt {x - 5} \\
  x - (x - 5) = \sqrt x + \sqrt {x - 5} \\
  x - x + 5 = \left( {\sqrt x + \sqrt {x - 5} } \right) \;
  $
 $ 5 = \sqrt x + \sqrt {x - 5} $ ------equation 2
Now we need to solve equation 1 and 2 for x ,
 $ \sqrt x - \sqrt {x - 5} = 1 $ equation 1
 $ 5 = \sqrt x + \sqrt {x - 5} $ equation 2
We added both the equations and $ \sqrt {x - 5} $ got cancelled out and we get the sum as ,
 $
\Rightarrow 2\sqrt x = 6 \\
\Rightarrow \sqrt x = \dfrac{6}{2} = 3 \\
\Rightarrow \sqrt x = 3 \;
  $
But we want the value of x , so we will do squaring both the sides ,
 $\Rightarrow x = 9 $
This is the required solution and our final answer .

Note: We can use prime factorisation for the number inside the radical and pull out non- radical terms or perfect squares from the inside of the square root to make the solution easier .
: In equivalent equations which have identical solutions we can perform multiplication or division by the same non-zero number both L.H.S. and R.H.S. of an equation .
In an equivalent equation which has an identical solution we can raise the same odd power to both L.H.S. and R.H.S. of an equation .
Cross check the answer and always keep the final answer simplified .