Problem 64 - Elementary Algebra

Question

For the following problems, round each result to the nearest hundredth. \(6 m^{2}-5=0\)

Step-by-Step Solution

Answer

Question: Solve the quadratic equation \(6 m^{2}-5=0\) and round the answer to the nearest hundredth. Answer: \(m \approx -0.29\) and \(m \approx 2.89\)

Step 1: Identify the coefficients

Find the coefficients in the given equation \(6 m^{2}-5=0\). We have: - \(a = 6\) - \(b = 0\) - \(c = -5\)

Step 2: Apply the quadratic formula

Use the quadratic formula to solve for m: \(m = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) Plug in the values for a, b, and c: \(m = \frac{-0 \pm \sqrt{0^2-4(6)(-5)}}{2(6)}\)

Step 3: Simplify and compute

Now we need to simplify the expression, and then use a calculator to find the final result: \(m = \frac{\pm \sqrt{-(-120)}}{12} = \frac{\pm \sqrt{120}}{12}\) Using a calculator, we find: - \(m \approx -\frac{\sqrt{120}}{12} \approx -0.29\) - \(m \approx \frac{\sqrt{120}}{12} \approx 2.89\) Thus, the rounded solutions to the nearest hundredth for the given equation \(6 m^{2}-5=0\) are \(m \approx -0.29\) and \(m \approx 2.89\).

Key Concepts

Rounding Numbers

Rounding is a technique used to simplify numbers, making them easier to work with. When rounding to the nearest hundredth, we look at the third place after the decimal point, known as the "thousandths place." Here’s how it works:

• If the thousandths digit is 5 or more, then we increase the hundredths digit by 1.
• If the thousandths digit is less than 5, we leave the hundredths digit unchanged.

For example, when you have a number like 2.891, you observe the digit 1 in the thousandths place.
Since 1 is less than 5, the number remains 2.89 when rounded to the nearest hundredth. Conversely, if we had 2.896, the digit 6 in the thousandths place would bump 9 in the hundredths place up to 10, making it 2.90. Understanding rounding helps ensure your math calculations remain precise and clear.

Quadratic Formula

The quadratic formula is a universal method used to solve quadratic equations, which are equations in the form of \( ax^2 + bx + c = 0 \).
It’s particularly useful when factoring is not possible. The formula is:\[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula helps find the roots (solutions) of the quadratic equation, where \( a \), \( b \), and \( c \) are coefficients from the equation.
Using the quadratic formula involves three main steps:

• Identify the coefficients: Recognize \( a \), \( b \), and \( c \) in your given equation.
• Calculate the discriminant: This is the part under the square root, \( b^2 - 4ac \), which determines the nature of the roots.
• Solve for the variable: Plug these values into the quadratic formula to compute the solutions.

This formula will give you two solutions, indicating where the parabola represented by the quadratic equation will intersect the x-axis.

Algebraic Solutions

Algebraic solutions involve using algebraic methods to find unknown values in equations. When tackling quadratic equations, as seen in the exercise:\( 6m^2 - 5 = 0 \), algebraic solutions connect well with real-world problems, like predicting the trajectory of an object.
Solving algebraically with the quadratic formula achieves precise results. Once the formula is applied and simplified, using techniques like squaring and extracting square roots, students are able to solve for the unknown variable.
This process frequently involves manipulating the equation through substitution or elimination. It’s crucial to follow a step-by-step method to ensure clarity and accuracy:

• Substitute the values of \(a\), \(b\), and \(c\) appropriately in the quadratic formula.
• Work through any arithmetic operations, being cautious with signs and operations.
• Simplify the expressions, and, if necessary, round your answers to the desired precision.

Algebraic solutions not only solve the equation but provide deeper insight into mathematical relationships and understanding.