Question

How do you solve \[3x\left( {x - 5} \right) = 0\] ?

Answer 1

Hint: This question involves the arithmetic operations like addition/ subtraction/ multiplication/ division. We need to know how to multiply the variable with the constant term, the constant term with the constant term, and the variable term with the variable term to solve these types of problems. Also, we need to know the basic algebraic formula to solve these types of questions.

Complete step by step solution:
The given equation is shown below,
 \[3x\left( {x - 5} \right) = 0 \to \left( 1 \right)\]
We have two-term in the above equation.
 First-term is \[3x\] and the second term is \[x - 5\] .
On the RHS side, we have \[0\] . So, the equation \[\left( 1 \right)\] has two solutions in two cases.
Case: \[1\]
Let’s move the term \[x - 5\] from the left-hand side to the right-hand side of the equation \[\left( 1 \right)\] , we get
 \[\left( 1 \right) \to 3x\left( {x - 5} \right) = 0\]
 \[3x = \dfrac{0}{{x - 5}}\]
We know that \[0\] divided by anything becomes zero. So, we get
 \[3x = 0\]
Let’s move the term \[3\] from the left-hand side to the right-hand side of the above equation, we get
 \[x = \dfrac{0}{3}\]
 \[x = 0\]
Case: \[2\] \[
    \\
    \\
 \]
Let’s move the term \[3x\] from the left-hand side to the right-hand side of the equation \[\left( 1 \right)\] , we get
 \[\left( 1 \right) \to 3x\left( {x - 5} \right) = 0\]
 \[x - 5 = \dfrac{0}{{3x}}\]
We know that \[0\] divided by anything the answer becomes \[0\] . So, we get
 \[x - 5 = 0\]
Let’s move the term \[ - 5\] from the left-hand side to the right-hand side of the above equation, we get
 \[x = 5\]
So, the final answer is,
 \[x = 0\] Or \[x = 5\]
So, the correct answer is “ \[x = 0\] Or \[x = 5\] ”.

Note: This question describes the arithmetic operations like addition/ subtraction/ multiplication/ division. This problem can also be solved by using the quadratic formula \[\left( {3x\left( {x - 5} \right) = 3{x^2} - 15x} \right)\] . Note that zero divided by anything becomes zero. Also, note that when we move one term from LHS to RHS or RHS to LHS, the arithmetic operations can be modified as given below,
I.Addition \[ \to \] subtraction
II.Subtraction \[ \to \] addition
III.Multiplication \[ \to \] division
IV.division \[ \to \] multiplication