Question

How do you solve the given equation $ {{\left( x-6 \right)}^{2}}=49 $ ?

Answer 1

Hint: We start solving the problem by expanding the square that is present in the L.H.S (Left Hand Side) of the given equation using the fact that $ {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab $ . We then make the necessary calculations and factorize the obtained equation. We then equate the factors obtained after factorization to 0 to get the required solution(s) of the given equation.

Complete step by step answer:
According to the problem, we are asked to solve the given equation $ {{\left( x-6 \right)}^{2}}=49 $ .
We have the equation $ {{\left( x-6 \right)}^{2}}=49 $ ---(1).
We know that $ {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab $ . Let us use this result in equation (1).
 $ \Rightarrow {{x}^{2}}+{{6}^{2}}-2\left( x \right)\left( 6 \right)=49 $ .
 $ \Rightarrow {{x}^{2}}-12x+36-49=0 $ .
 $ \Rightarrow {{x}^{2}}-12x-13=0 $ .
Now, let us factorize the obtained equation to get the required solution.
 $ \Rightarrow {{x}^{2}}-13x+x-13=0 $ .
 $ \Rightarrow x\left( x-13 \right)+1\left( x-13 \right)=0 $ .
 $ \Rightarrow \left( x+1 \right)\left( x-13 \right)=0 $ ---(2).
Now, let us equate each of the obtained factors to 0 to get the solution for the quadratic equation.
 $ \Rightarrow x+1=0 $ , $ x-13=0 $ .
 $ \Rightarrow x=-1 $ , $ x=13 $ .
So, we have found the solution(s) for the given equation $ {{\left( x-6 \right)}^{2}}=49 $ as $ x=-1 $ and $ x=13 $ .
 $ \therefore $ The solution(s) for the given equation $ {{\left( x-6 \right)}^{2}}=49 $ is (are) $ x=-1 $ and $ x=13 $ .

Note:
Whenever we get this type of problem, we first try to factorize the given equation which makes the process easier to get a solution. We should not make calculation mistakes while solving this problem. We can also solve the given problems as shown below:
We have given the equation $ {{\left( x-6 \right)}^{2}}=49 $ .
 $ \Rightarrow {{\left( x-6 \right)}^{2}}={{7}^{2}} $ ---(3).
We know that if $ {{a}^{2}}={{b}^{2}} $ , then $ a=\pm b $ . Let us use this result in equation (3).
 $ \Rightarrow x-6=\pm 7 $ .
 $ \Rightarrow x-6=7 $ , $ x-6=-7 $ .
 $ \Rightarrow x=7+6 $ , $ x=-7+6 $ .
 $ \Rightarrow x=13 $ , $ x=-1 $ .
So, the solution(s) for the given equation $ {{\left( x-6 \right)}^{2}}=49 $ is (are) $ x=-1 $ and $ x=13 $ .