Question

How do you solve $ \sqrt {x - 4} + \sqrt {x + 1} = 5 $ ?

Answer 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 . After eliminating the radical , we will then solve the equation by applying the identity and find the value of the x .

Complete step-by-step answer:
If we see the question , we need to solve the given expression under square root which is $ \sqrt {x - 4} + \sqrt {x + 1} = 5 $ . -----equation 1
By applying the concept of equivalent equation , we will first do squaring both sides on $ \sqrt {x - 4} + \sqrt {x + 1} = 5 $ both the L . H . S . and the R . H . S . as follows –
We are going to isolate a square root on the L . H . S . So that we can apply the identity as stated below –
We can apply the formula or we can say identity $ {(a - b)^2} = {a^2} + {b^2} - 2ab $
 $ \sqrt {x - 4} + \sqrt {x + 1} = 5 $
 $ \sqrt {x - 4} = 5 - \sqrt {x + 1} $
 $ {\left( {\sqrt {x - 4} } \right)^2} = {\left( {5 - \sqrt {x + 1} } \right)^2} $
 $
\Rightarrow x - 4 = {5^2} - 10\sqrt {x + 1} + (x + 1) \\
\Rightarrow x - (x + 1) + 10\sqrt {x + 1} = 25 + 4 \\
   - 1 + 10\sqrt {x + 1} = 29 \\
  \sqrt {x + 1} = \dfrac{{30}}{{10}} \\
  \sqrt {x + 1} = 3 \;
  $
Now , squaring both the sides to get the original value of x , we get –
 $
\Rightarrow x + 1 = 9 \\
  x = 8 \;
  $
This is our required and final answer as x=8 .
So, the correct answer is “x=8”.

Note: Always remember the algebraic identities .
Always try to get rid of the square root .
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 .