How do you solve \[(x - 2)(x - 8) = 0\]?

Hint:The equation is an algebraic equation and here in this question we have to solve this equation. By transforming or by shifting the terms we have to find the value for the variable x. Since the equation is in the form of factors which are equated to zero, we can easily find the value of x.

Complete step-by-step solution:
The equation is an algebraic equation or expression. It is a combination of variables and constants. In algebraic equations we have arithmetic operations also. Here in this question, we have to find the value of x.
Here we have two terms and the two terms are in the braces. if the terms are in the braces then they are in the form of multiplication.
Now consider the equation \[(x - 2)(x - 8) = 0\]. since it is in the form of product either any one will be zero. if the first term is zero means we have
\[ \Rightarrow (x - 2) = 0\]
take 2 to RHS we get
\[ \Rightarrow x = 2\]
Suppose if the second term is zero means we have
\[ \Rightarrow (x - 8) = 0\]
take 8 to RHS we get
\[ \Rightarrow x = 8\]
Hence the x takes the value either 2 or 8.
We can also check the obtained values is correct or not by substituting these values in the given equation
consider \[(x - 2)(x - 8) = 0\]
Substitute the value of x has 2 we have
\[ \Rightarrow (2 - 2)(x - 8) = 0\]
\[ \Rightarrow 0(x - 8) = 0\]
Anything multiplied by the zero the solution will be zero so we have
\[ \Rightarrow 0 = 0\]
LHS = RHS
Substitute the value of x has 8 to the given equation we have
\[ \Rightarrow (x - 2)(8 - 8) = 0\]
\[ \Rightarrow (x - 2)0 = 0\]
Anything multiplied by the zero the solution will be zero so we have
\[ \Rightarrow 0 = 0\]
LHS = RHS
Therefore, the values of x are correct.
Hence, we have solved and obtained the solution. x = 2 or x = 8

Note: While solving the equation we shift or transform the terms either from LHS to RHS or from RHS to LHS we should take care of the sign. Because while shifting or transforming the terms the sign of the term will change. If we miss out the sign we may go wrong while finding the variable or solving.