Problem 46 - Algebra 1: Concepts and Skills

Question

Use the quadratic formula to solve the equation. If the solution involves radicals, round to the nearest hundredth. $$8 m^{2}+6 m-1=0$$

Step-by-Step Solution

Answer

The solutions to the equation are \(m = 0.18\) and \(m = -0.68\) when rounded to the nearest hundredth.

Step 1: Identify the coefficients

Let's identify the coefficients in the equation \(8m^2 + 6m - 1 = 0\). Here \(a = 8\), \(b = 6\), and \(c = -1\).

Step 2: Substitute the coefficients into the quadratic formula

Substitute the coefficients into the quadratic formula which is \(-b \pm \sqrt{b^2 - 4ac}\) over \(2a\). So, the solution to the equation is given by: \[m = \frac{-6 \pm \sqrt{(6)^2 - 4*8*(-1)}}{2*8}\]. Simplify the expression inside the square root first.

Step 3: Simplify the expression

Now, compute the expression inside the square root: \( (6)^2 - 4*8*(-1)=36 + 32 = 68\). So the equation becomes \[m = \frac{-6 \pm \sqrt{68}}{16}\].

Step 4: Calculate the solutions

Now, separate it into two equations and calculate the solutions: \[m1 = \frac{-6 + \sqrt{68}}{16} , m2 = \frac{-6 - \sqrt{68}}{16}\]. Solve these to get the roots of the quadratic equation. After rounding to the nearest hundredth, the roots are \(m1 = 0.18\) and \(m2 = -0.68\).

Key Concepts

Solving Quadratic Equations

When faced with a quadratic equation, such as the given textbook exercise \( 8m^2 + 6m - 1 = 0 \), the goal is to find the values of the variable that make the equation true. These values are known as the \(roots\text{ or }\)\(x-intercepts\) of the equation.
A quadratic equation is any equation that can be written in the form \( ax^2 + bx + c = 0 \), where \(a\text{, }\)\(b\text{, and }\)\(c\) are constants, and \(a \eq 0\text{.}\) One of the most powerful tools for solving these equations is the quadratic formula, which provides a straightforward method for finding the roots.
Improving upon a student's understanding of how to solve quadratic equations can start with reinforcing the process of \(identifying the coefficients\) – in this case, \(a = 8\text{, }\)\(b = 6\text{, and }\)\(c = -1\). From there, the key is practicing substituting these values into the quadratic formula and simplifying the resulting expression, as shown in step 2 and step 3 of the provided step-by-step solution.
Regularly attempting various quadratic equations with differing coefficients helps to solidify the understanding of this concept and improve the ability to solve them quickly and accurately.

Quadratic Equation Roots

The roots of a quadratic equation, or the solutions to the equation, are the values of \(x\) that satisfy the equation \(ax^2 + bx + c = 0\text{.}\) In the exercise, we used the quadratic formula to calculate the roots of \(8m^2 + 6m - 1 = 0\text{.}\) The roots, \(m1\text{ and }\)\(m2\text{,}\) indicate the points where the parabola described by the quadratic equation crosses the x-axis.
Understanding the concept of roots is crucial as it emphasizes the reason behind learning to solve a quadratic equation: finding where the graph of the equation touches or crosses the x-axis. These roots can be real or complex numbers and can be distinct, coincident, or non-real depending on the discriminant \(b^2 - 4ac\text{.}\)
To aid comprehension, identifying that the quadratic formula can yield two solutions—represented by the \(\pm\) sign in the numerator—is important. These correspond to the two possible points where the equation intersects the x-axis, as illustrated in step 4 of the solution. Relating the concept of roots to graphical interpretations can provide students with a visual understanding of the roots.

Radicals in Algebra

In algebra, radicals represent the roots of numbers and expressions, and they often appear when working with quadratic equations, especially in the quadratic formula. In our exercise, we encounter the square root of 68, represented as \( \sqrt{68} \text{,}\) which is a radical.
The presence of a radical, especially a square root, indicates that you are dealing with the root of a number. Understanding how to simplify radicals and handle them in calculations is a fundamental skill in algebra. For instance, simplifying \( \sqrt{68} \text{,}\) may involve expressing it as \( \sqrt{4 \cdot 17} \text{,}\) which then simplifies further to \( 2\sqrt{17} \text{,}\) thus making the numerical computation and interpretation simpler.
Furthermore, when students are instructed to round radicals to the nearest hundredth, as seen in the solution, it highlights the real-life application of radicals where exact values are not always necessary, and approximate values can suffice. Always consider the context to decide whether an exact radical form or a decimal approximation is more appropriate. Exercises involving radicals sharpen estimations skills and enhance number sense in algebraic contexts.