Problem 67

Question

For the following problems, solve the equations, if possible. $$ 3 m n^{2}-36 m n+36 m=0 $$

Step-by-Step Solution

Verified
Answer
Question: Solve the equation \(3mn^2-36mn+36m=0\) and find the values of 'm' and 'n'. Answer: The solutions to the given equation are \(m = 0\) and \(n = 6\).
1Step 1: Identify the common factor
Identify the common factor in all the terms of the given equation, which is 'm'.
2Step 2: Factor out the common factor
Factor 'm' out of each term in the expression: $$ m(3n^2 - 36n + 36) = 0 $$
3Step 3: Simplify the expression inside the parentheses
Simplify the expression inside the parentheses by finding the common factor for the terms within: $$ 3(n^2 - 12n + 12) = 0 $$
4Step 4: Factor the quadratic expression
Now, we need to factor the quadratic expression \((n^2 - 12n + 12)\): $$ 3(n-6)^2 = 0 $$
5Step 5: Set each factor to zero
The equation is now factored and can be solved by setting each factor to zero: $$ m=0 \quad or\quad (n-6)^2=0 $$
6Step 6: Solve for each variable
Now, we solve for 'm' and 'n': $$ m = 0 $$ and $$ (n-6)^2 = 0 \Rightarrow n = 6 $$ So, the solutions are \(m = 0\) and \(n = 6\).

Key Concepts

Factoring PolynomialsCommon Factor in AlgebraQuadratic ExpressionAlgebraic Equation Solutions
Factoring Polynomials
Factoring polynomials is a crucial technique in algebra that simplifies equations and makes it easier to find solutions. To factor a polynomial, you need to express it as a product of simpler polynomials. This can be thought of as breaking down a large composite number into its prime factors. By simplifying a polynomial, you can work with smaller, more manageable terms that are easier to solve.

For instance, in the given problem, we have an expression:
  • Original equation: \(3mn^2 - 36mn + 36m = 0\)
  • First identify the common factor: \(m\)
  • Result after initial factoring: \(m(3n^2 - 36n + 36) = 0\)
In this way, we recognize a common factor across each term and factor it out to ease the complexity of the expression.
Common Factor in Algebra
Finding the common factor is an important step in simplifying algebraic equations. A common factor is a term that appears in every part of the expression. Identifying it helps in reducing the scale of the numbers you work with in calculations, making it easier to manipulate equations.

In our case, the equation \(3mn^2 - 36mn + 36m = 0\) has a common factor of \(m\) in each term.
  • First term: \(3mn^2\) contains \(m\)
  • Second term: \(-36mn\) contains \(m\)
  • Third term: \(36m\) contains \(m\)
Once you figure out that \(m\) is present in all terms, you can factor it out:\( m(3n^2 - 36n + 36) = 0 \)
This simplifies the process of handling and solving the equation.
Quadratic Expression
A quadratic expression is a polynomial of degree 2, typically taking the form \(ax^2 + bx + c\). It represents a parabola when graphed. Solving quadratic expressions often requires factoring, using the quadratic formula, or completing the square.
In this exercise, the reduced equation inside the parentheses is quadratic: \(n^2 - 12n + 12\).
  • Start by identifying the leading term, \(n^2\)
  • Notice the reduction process: \(3(n^2 - 12n + 12) = 0\)
After simplification, this quadratic equation represented a perfect square trinomial:\( (n-6)^2 = 0 \)
This shows that there is one real repeated solution for the variable \(n\). Understanding quadratic expressions helps unravel the solutions for such equations.
Algebraic Equation Solutions
Solving algebraic equations involves finding the value of variables that make the equation true. In linear equations, there's typically one solution, whereas quadratic equations can have up to two solutions.
For this equation, begin by setting each factor to zero, which is a key strategy for solving equations after factoring:
From the factors of the equation \(m(3(n-6)^2) = 0\), the possible solutions are:
  • \(m = 0\)
  • \((n-6)^2 = 0 \Rightarrow n = 6\)
By isolating each factor and solving for the variables, you find the solutions \(m = 0\) and \(n = 6\). This step-by-step approach enables a clearer pathway to deriving the values that satisfy the original equation.