Question

How do you solve $ \sqrt {2x + 7} = 6 $ ?

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 simplifying and finding 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 {2x + 7} = 6 $ . -----equation 1
By applying the concept of equivalent equation , we will first do squaring both sides on $ \sqrt {2x + 7} = 6 $ 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 simplifying as stated below –
 $ \sqrt {2x + 7} = 6 $
 $ {\left( {\sqrt {2x + 7} } \right)^2} = {\left( 6 \right)^2} $
 $ 2x + 7 = 36 $
Now subtract seven from both the L . H . S . and the R . H . S . , we get
 $\Rightarrow 2x = 36 - 7 $
 $\Rightarrow 2x = 29 $
Dividing both the sides by 2 , we get –
 $\Rightarrow x = \dfrac{{29}}{2} $
Therefore , the required and the final answer is $ x = \dfrac{{29}}{2} $ .
So, the correct answer is “ $ x = \dfrac{{29}}{2} $ ”.

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 .