Question

How do you solve $ \sqrt {3x + 4} + x = 8 $ and find any extraneous solution ?

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 square root which is $ \sqrt {3x + 4} + x = 8 $ . -----equation 1
By applying the concept of equivalent equation , we will first do squaring both sides on $ \sqrt {3x + 4} + x = 8 $ 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 $
Before all this Subtract the x from both the sides L . H . S . and the R . H . S .
 $ \sqrt {3x + 4} + x = 8 $
 $ \sqrt {3x + 4} = 8 - x $
 $ {\left( {\sqrt {3x + 4} } \right)^2} = {\left( {8 - x} \right)^2} $
 $ 3x + 4 = 64 - 16x + {x^2} $
Subtract $ 3x + 4 $ from both the sides L . H . S . and the R . H . S. , we get –
 $ 0 = {x^2} - 19x + 60 $
We will take out the factors from the equation ,
 $
\Rightarrow (x - 4)(x - 15) = 0 \\
\Rightarrow x = 4,x = 15 \;
  $
We can check for an extraneous solution that means which is not the actual solution of an original equation .
Substitute 4 and 15 in the original equation and check if the L . H . S . and the R . H . S. becomes equal ,
Putting 4 , we get –
 $ \sqrt {3(4) + 4} + 4 = 8 $
8=8
Hence , verified .
Putting 15 , we get –
 $ \sqrt {3(15) + 4} + 15 = 8 $
22=8
This does not satisfy .
Therefore $ x = 4 $ is the original solution and the value $ x = 15 $ is the extraneous solution .
The final answer is $ x = 4 $ .
So, the correct answer is “ $ x = 4 $ ”.

Note: 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 .