Question

How do you solve \[| - 4x|\, = 32\] ?

Answer 1

Hint: Here the equation is an algebraic equation that is a combination of constant and variables. we have to solve the given equation for variable x. Since the equation involves the modulus, by using the definition of modulus or absolute value and simple arithmetic operation we determine the value of x.

Complete step by step solution:
The absolute value or modulus of a real function f(x), it is denoted as |f(x)|, is the non-negative value of f(x) without considering its sign. \[| - 4x|\, = 32\]
By the definition of absolute number we are determined the unknown variable and
By definition the modulus, separate \[| - 4x|\, = 32\] into two equations:
 \[ - 4x = 32\] (1)
 and
 \[ - ( - 4x) = 32\] (2)
Consider the equation (1)
 \[ \Rightarrow - 4x = 32\]
Divide the above equation by -4
 \[\therefore x = - 8\]
Now consider the equation (2)
 \[ \Rightarrow - ( - 4x) = 32\]
First multiply the -ve sign inside to the parenthesis on LHS.
 \[ \Rightarrow 4x = 32\]
Divide the above equation by 4
 \[\therefore x = 8\]
Hence, the value of x in the equation \[| - 4x|\, = 32\] is -8 and 8.
Now we will verify the obtained solutions
Verification:
Put x=-8 to the equation \[| - 4x|\,\] , then
 \[ \Rightarrow |4( - 8)|\,\]
On multiplying we have
 \[ \Rightarrow | - 32|\,\]
On simplification we have
 \[ \Rightarrow 32\]
 \[\therefore | - 4x|\, = 32\]
Put x=8 to the equation \[| - 4x|\,\] , then
 \[ \Rightarrow |4(8)|\,\]
On multiplying we have
 \[ \Rightarrow |32|\,\]
On simplification we have
 \[ \Rightarrow 32\]
 \[\therefore | - 4x|\, = 32\]
Hence verified.
Since the RHS of the given equation is zero we obtain only one solution.

Note: The algebraic equation or an expression is a combination of variables and constants, it also contains the coefficient. The alphabets are known as variables. The x, y, z etc., are called as variables. The numerals are known as constants. The numeral of a variable is known as co-efficient. we must know about the modulus definition.