How do you solve \[{{w}^{2}}=4\] where \[w\] is a real number?

Hint:As we have given a quadratic equation with the coefficient of a linear variable is zero. First of all, make this equation of type equal to zero that is \[{{w}^{2}}-4=0\] then use identity \[{{a}^{2}}-{{b}^{2}}=(a-b)(a+b)\] and comparing the given question with this identity and the RHS will remain zero as we used this identity in LHS. Now the product of two terms is zero then it means either of the terms is zero or both the terms are zero then equate with you will get the answer.

Complete Solution:
First of all, make this equation of type equal to zero
\[\Rightarrow {{w}^{2}}-4=0\]
Since the given equation is a quadratic equation with the coefficient of the linear variable is zero.
As we have only quadratic and the constant term,
So, we will use identity \[{{a}^{2}}-{{b}^{2}}=(a-b)(a+b)\]
Now comparing this identity with the above question, we get
\[{{a}^{2}}={{w}^{2}}\] and \[{{b}^{2}}=4\]
\[\Rightarrow {{w}^{2}}-4=0\]
\[\Rightarrow (w-2)(w+2)=0\]
Here the product of two terms is zero that means either of term is zero or both are zero
\[\Rightarrow w-2=0\] or \[w+2=0\]
On simplifying, we get
\[w=2\] or \[w=-2\]
Also given that \[w\] is a real number and that is just verified

Hence \[w=2\] or \[w=-2\]

Note: In this type of quadratic equations where the coefficient of a linear variable is zero we can apply the above identity or from the question we can clearly see that it is given that the square of an unknown variable is \[4\] then we can take square root both sides and a plus-minus both sign will come before square root and after square root, we have directly calculated the variable \[w\].