Problem 36 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Solve the equation by using the quadratic formula. $$ 0.2 m^{2}+1.6 m+1.2=0 $$

Step-by-Step Solution

Answer

Using the quadratic formula and plugging in the values \(a = 0.2\), \(b = 1.6\), and \(c = 1.2\), we get the solutions for the given equation as \(m_1 \approx -1.23265\) and \(m_2 \approx -4.76735\).

Step 1: Write down the quadratic formula

The quadratic formula is given as follows: \[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Step 2: Identify the values of a, b, and c

From the given equation, \(0.2m^2 + 1.6m + 1.2 = 0\), we can identify the values as: \[a = 0.2, \quad b = 1.6, \quad c = 1.2\]

Step 3: Plug in the values of a, b, and c into the quadratic formula

Now, substitute the values of \(a, b,\) and \(c\) in the quadratic formula: \[m = \frac{-1.6 \pm \sqrt{(1.6)^2 - 4 \cdot 0.2 \cdot 1.2}}{2 \cdot 0.2}\]

Step 4: Simplify the expression under the square root

We need to simplify the expression inside the square root: \[\sqrt{(1.6)^2 - 4 \cdot 0.2 \cdot 1.2} = \sqrt{2.56 - 0.96} = \sqrt{1.6}\]

Step 5: Simplify the expression

Now, substitute the simplified square root back into the formula: \[m = \frac{-1.6 \pm \sqrt{1.6}}{2 \cdot 0.2}\]

Step 6: Calculate the two possible values of m

We need to calculate the values of \(m\) using the plus and the minus sign: \[m_1 = \frac{-1.6 + \sqrt{1.6}}{0.4}\] \[m_2 = \frac{-1.6 - \sqrt{1.6}}{0.4}\] Calculate these values using a calculator to get: \[m_1 \approx -1.23265\] \[m_2 \approx -4.76735\]

Step 7: Write down the solutions

The solutions for the given quadratic equation are: \[m_1 \approx -1.23265\] \[m_2 \approx -4.76735\]

Key Concepts

Quadratic Formula

Solving quadratic equations is a fundamental skill in algebra, and the quadratic formula is one of the most powerful tools available for this purpose. Quadratic equations often appear in the form of \[ ax^2 + bx + c = 0 \], where \( a \), \( b \), and \( c \) are constants. The quadratic formula provides a way to find the solutions for such equations. It is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula will provide all possible values of \( x \) that satisfy the equation. The "\( \pm \)" symbol indicates that there are generally two solutions, corresponding to the plus and the minus in the formula.Here's a helpful tip:

• First, clearly identify the values of \( a \), \( b \), and \( c \) from the equation.
• Then, substitute these values into the formula to find the solutions.

The quadratic formula is universally applicable to any quadratic equation, making it an essential part of solving algebraic problems.

Mathematical Problem Solving

Using the quadratic formula is a prime example of mathematical problem solving. Here's how you can apply the formula to tackle problems like the one you've just seen:

• Understand the Problem: Start by writing down the quadratic equation and identify its coefficients \( a \), \( b \), and \( c \). In our example, \( a = 0.2 \), \( b = 1.6 \), and \( c = 1.2 \).
• Solve the Equation: Plug these values into the quadratic formula. Calculate the discriminant \( b^2 - 4ac \) first, as it will determine the nature of the solutions.
• Simplify and Calculate: Once you have the discriminant, simplify the expression under the square root. Continue by implementing the plus and minus operations to find two possible solutions.

This process not only provides a solution to the problem at hand but also develops critical thinking and analytical skills. Each step should be followed carefully, as errors often occur during calculation and simplification.

Algebra Concepts

Quadratic equations are a significant part of algebra concepts. They feature prominently in various real-world applications, from physics and engineering to finance and statistics. Understanding how to work with these equations provides a strong foundation for tackling more complex mathematical challenges. Key algebraic concepts linked to quadratic equations include:

• Factorization: Attempting to rewrite the equation in a factorable form can often provide insight into its solutions.
• Completing the Square: This method involves altering the equation to enable the direct application of the quadratic formula.
• Graphical Interpretations: A quadratic equation can be represented graphically as a parabola, with solutions corresponding to the points where the curve intersects the x-axis.

Grasping these concepts is crucial for solving algebraic problems effectively and is a vital skill for any student studying mathematics.