Question

How do you solve $ \sqrt {x + 4} = 1 - \sqrt {3x + 13} $ ?

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 , 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 sign . . 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 the square root which is $ \sqrt {x + 4} = 1 - \sqrt {3x + 13} $ . -----equation 1
By applying the concept of equivalent equation , we will first do squaring both sides on $ \sqrt {x + 4} = 1 - \sqrt {3x + 13} $ 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 $
 $
\Rightarrow x + 4 = 1 - 2\sqrt {3x + 13} + (3x + 13) \\
\Rightarrow x + 4 = (3x + 14) - 2\sqrt {3x + 13} \;
  $
Now we will add $ 2\sqrt {3x + 13} - x - 4 $ to both the sides L . H . S . and the R . H . S . as follows –
 $ 2\sqrt {3x + 13} = 2x + 10 $
Further dividing both the sides by 2 , we get –
 $ \sqrt {3x + 13} = x + 5 $
Again , squaring both the sides and applying the identity , we get –
 $ 3x + 13 = {x^2} + 10x + 5 $
Now subtract 3x + 13 from both sides to get a quadratic equation :
 $ 0 = {x^2} + 7x + 12 $
We can take their roots out by factorization ,
 $ \left( {x + 3} \right)\left( {x + 4} \right) $
So , it comes out to be ,
 $ x = - 1,x = - 4 $
Now we will put these roots to check for extraneous roots ,
 $
  \sqrt {( - 3) + 4} = 1 \ne - 1 = 1 - 2 = 1 - \sqrt 4 \ne 1 - \sqrt {3( - 3) + 13} \\
  \sqrt {( - 4) + 4} = 0 \ne 1 = 1 - 1 = 1 - \sqrt 1 = 1 - \sqrt {3( - 4) + 13} \\
  $
Therefore , $ x = - 3 $ is an extraneous solution that is not an actual solution of the original equation which is introduced by squaring $ 1 \ne - 1 $ while $ x = - 4 $ is the actual solution of the original equation .
The final answer is $ x = - 4 $ .
So, the correct answer is “ $ x = - 4 $ ”.

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 .