Chapter 1: Problem 36
Solve the equation by using the quadratic formula. $$ 0.2 m^{2}+1.6 m+1.2=0 $$
Short Answer
Expert verified
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 by step solution
01
Write down the quadratic formula
The quadratic formula is given as follows:
\[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
02
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:
\[a = 0.2, \quad b = 1.6, \quad c = 1.2\]
03
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:
\[m = \frac{-1.6 \pm \sqrt{(1.6)^2 - 4 \cdot 0.2 \cdot 1.2}}{2 \cdot 0.2}\]
04
Simplify the expression under the square root
We need to simplify the expression inside the square root:
\[\sqrt{(1.6)^2 - 4 \cdot 0.2 \cdot 1.2} = \sqrt{2.56 - 0.96} = \sqrt{1.6}\]
05
Simplify the expression
Now, substitute the simplified square root back into the formula:
\[m = \frac{-1.6 \pm \sqrt{1.6}}{2 \cdot 0.2}\]
06
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:
\[m_1 \approx -1.23265\]
\[m_2 \approx -4.76735\]
07
Write down the solutions
The solutions for the given quadratic equation are:
\[m_1 \approx -1.23265\]
\[m_2 \approx -4.76735\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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.