Problem 18 - Introductory Algebra for College Students

Question

Solve each equation using the quadratic formula. Simplify irrational solutions, if possible. $$3 x^{2}-5 x+1=0$$

Step-by-Step Solution

Answer

The solutions to the quadratic equation \(3x^{2}-5x+1=0\) are \(x = \frac{5+\sqrt{13}}{6}\) and \(x = \frac{5-\sqrt{13}}{6}\).

Step 1: Identify a, b, and c

The quadratic formula involves the coefficients a, b, and c from the quadratic equation in the form \(ax^{2}+bx+c=0\). Here, \(a = 3\), \(b = -5\), and \(c = 1\).

Step 2: Apply quadratic formula

Substitute the values of a, b, and c into the quadratic formula to find the solutions for x. So, \(x = \frac{-(-5) \pm \sqrt{(-5)^{2}-4*3*1}}{2*3} = \frac{5 \pm \sqrt{25-12}}{6}\).

Step 3: Simplify the determinant

The determinant is \(25-12 = 13\), which leaves us with \(x = \frac{5 \pm \sqrt{13}}{6}\).

Step 4: Find the solutions

The quadratic equation has two solutions, which can be found by doing the plus and minus calculations separately. They are \(x = \frac{5+\sqrt{13}}{6}\) and \(x = \frac{5-\sqrt{13}}{6}\).

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation involving a squared term, typically written in the standard form of \(ax^2 + bx + c = 0\). These equations are a fundamental part of algebra and are important for various fields and applications. In a quadratic equation, \(a\), \(b\), and \(c\) are numerical coefficients, and \(x\) represents the variable or unknown that we seek to solve.

Key characteristics of quadratic equations include:

• The highest power of the unknown \(x\) is 2, hence the name "quadratic" derived from "quad" meaning square.
• Quadratic equations always form a parabola when graphed, opening upwards if \(a > 0\) and downwards if \(a < 0\).
• They can have either two real solutions, one real solution, or two complex solutions depending on the value of the discriminant \(b^2 - 4ac\).

Understanding the structure of these equations helps in learning different methods for finding their solutions, including factoring, completing the square, and using the quadratic formula.

Solving Equations

Solving equations, particularly quadratic equations, entails finding the values of \(x\) that satisfy the equation. There are several methods to solve quadratic equations, but the quadratic formula is one of the most elegant and reliable.

The steps in using the quadratic formula include:

• Identifying the coefficients \(a\), \(b\), and \(c\) from the equation.
• Substituting these values into the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• Simplifying the expression under the square root, known as the discriminant.
• Performing the mathematical operations to arrive at the solution(s) for \(x\).

This method is especially useful when the equation is not easily factorable. The key challenge in this method often lies in correctly simplifying the expression under the square root and ensuring accurate arithmetic operations.
It is also essential to note that the discriminant \(b^2 - 4ac\) informs us about the number and nature of the roots of the equation:

• A positive discriminant means two distinct real roots.
• A zero discriminant means one real root (a repeated root).
• A negative discriminant results in two complex roots.

Algebraic Solutions

In algebra, solving equations involves manipulating algebraic expressions to isolate the variable. For quadratic equations, the algebraic solution process using the quadratic formula is systematic and can be applied to any quadratic equation.

To solve our equation, \(3x^2 - 5x + 1 = 0\), we identify the coefficients:

• \(a = 3\)
• \(b = -5\)
• \(c = 1\)

With these coefficients, substitute them into the quadratic formula: \[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} = \frac{5 \pm \sqrt{13}}{6} \]This results because the discriminant, \((-5)^2 - 4 \cdot 3 \cdot 1\), evaluates to 13, which is positive, indicating two real roots for the equation.

By performing the arithmetic, we arrive at two exact solutions:

• \(x = \frac{5 + \sqrt{13}}{6}\)
• \(x = \frac{5 - \sqrt{13}}{6}\)

These solutions demonstrate a complete algebraic approach in finding the values of \(x\) for the quadratic equation, showcasing the power of algebraic techniques in solving problems effectively.