Problem 25 - Algebra 2

Question

Solve each equation using the Quadratic Formula. Find the exact solutions. Then approximate any radical solutions. Round to the nearest hundredth. $$ 6 x^{2}-5 x-1=0 $$

Step-by-Step Solution

Answer

The exact solutions are \(x = 1\) and \(x = -\frac{1}{6}\). Since the solutions are rational numbers, no approximation is needed.

Step 1: Identifying coefficients

Identify the coefficients of the quadratic equation. In the equation \(6x^{2} - 5x - 1 = 0\), the coefficient for \(x^{2}\) is 6 (\(a = 6\)), the coefficient for \(x\) is -5 (\(b = -5\)), and the constant term is -1 (\(c = -1\)).

Step 2: Applying the Quadratic Formula

Apply the quadratic formula to find the roots of the equation. The quadratic formula is \(x = \frac{{-b \pm \sqrt{{b^{2} - 4ac}}}}{{2a}}\). Substituting the identified coefficients gives us \(x = \frac{{-(-5) \pm \sqrt{{(-5)^{2} - 4(6)(-1)}}}}{{2(6)}}\).

Step 3: Calculating the discriminant

Calculate the discriminant, which is the part under the square root in the quadratic formula: \(b^{2} - 4ac\). Compute the value of the discriminant: \((-5)^{2} - 4(6)(-1) = 25 + 24 = 49\).

Step 4: Finding the exact roots

With the discriminant calculated, find the exact roots by continuing with the formula: \(x = \frac{{5 \pm \sqrt{{49}}}}{{12}}\). Since \(\sqrt{49} = 7\), we have \(x = \frac{{5 \pm 7}}{{12}}\).

Step 5: Simplifying the solutions

Simplify the two possible solutions: \(x = \frac{{5 + 7}}{{12}} = \frac{{12}}{{12}} = 1\) and \(x = \frac{{5 - 7}}{{12}} = \frac{{-2}}{{12}} = -\frac{1}{6}\).

Step 6: Approximating radical solutions

Since both solutions are rational numbers, there are no radicals to approximate. If there were, we would round them to the nearest hundredth.

Key Concepts

Solving Quadratic Equations

Quadratic equations are a foundational concept in algebra, characterized by an equation in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are known values and \(a \eq 0\). The Quadratic Formula is a powerful tool for solving such equations, expressed as \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\).

To apply the formula effectively, first identify the coefficients \(a\), \(b\), and \(c\) from the equation. Next, calculate the value under the square root, known as the discriminant, which determines the nature of the roots. Finally, solve for \(x\) using the plus or minus sign to find both possible roots. For example:

• Identify coefficients in \(6x^2 - 5x - 1 = 0\): \(a = 6\), \(b = -5\), \(c = -1\).
• Apply the Quadratic Formula with these coefficients.
• Simplify the expressions to find the exact solutions.

The resulting roots will either be real and distinct, real and repeated, or complex, depending on the discriminant's value.

Discriminant in Quadratics

The discriminant in a quadratic equation is the part of the Quadratic Formula under the radical sign: \({b^2 - 4ac}\). It plays a critical role in determining the quantity and type of solutions.

The discriminant can reveal:

• If it's positive, there are two distinct real roots.
• If it's zero, there is one real root (also called a repeated or double root).
• If it's negative, there are two complex roots, which are not real numbers.

In the example \(6x^2 - 5x - 1 = 0\), the computed discriminant is \({(-5)^2 - 4(6)(-1) = 49}\), which is positive. This tells us there are two distinct real solutions. Understanding the discriminant gives insight into the nature of roots without fully solving the equation, making it a valuable preliminary step in the process.

Approximating Radical Solutions

When solving quadratic equations, we sometimes encounter radical solutions—roots that include a square root term. If the discriminant is not a perfect square, the roots will be irrational, meaning they cannot be expressed as simple fractions.

In such cases, we approximate the radical solutions to a desired degree of accuracy, typically rounding them to the nearest hundredth. To approximate:

• First, find the decimal form of the radical expression.
• Then, round this decimal to the required precision.

This step is vital when exact decimal solutions are required, for instance, in applied mathematics or real-world measurements. For our example equation, \(6x^2 - 5x - 1 = 0\), the solutions were rational and did not require approximation. However, if the discriminant had been, say, \({48}\), we would have approximated \(x\) values to the nearest hundredth as the square root of \({48}\) doesn’t yield a neat fraction.