Problem 32 - Elementary Algebra

Question

Use the quadratic formula to solve the following. $$4 t 2-8 t-1=0$$

Step-by-Step Solution

Answer

The solutions are \(t = 1 + \frac{\sqrt{5}}{2}\) and \(t = 1 - \frac{\sqrt{5}}{2}\).

Step 1: Identifying Parameters

In a quadratic equation of the form \(ax^2 + bx + c = 0\), identify the coefficients \(a\), \(b\), and \(c\). For the equation \(4t^2 - 8t - 1 = 0\), we have:- \(a = 4\)- \(b = -8\)- \(c = -1\)

Step 2: Writing the Quadratic Formula

The quadratic formula to find the roots of \(ax^2 + bx + c = 0\) is given by:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Step 3: Substitute the Parameters

Substitute the values of \(a\), \(b\), and \(c\) into the formula:\[ t = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \times 4 \times (-1)}}{2 \times 4} \]

Step 4: Simplifying the Expression

Simplify the expression inside the square root:\[ t = \frac{8 \pm \sqrt{64 + 16}}{8} \]\[ t = \frac{8 \pm \sqrt{80}}{8} \]

Step 5: Solve the Square Root

Calculate \( \sqrt{80} \):\[ \sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5} \]Substitute back into the equation:\[ t = \frac{8 \pm 4\sqrt{5}}{8} \]

Step 6: Final Simplification

Simplify the division of each term by 8:\[ t = 1 \pm \frac{\sqrt{5}}{2} \]So, the solutions are:\[ t_1 = 1 + \frac{\sqrt{5}}{2} \]\[ t_2 = 1 - \frac{\sqrt{5}}{2} \]

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation of degree 2, which means its highest power is squared. The standard form for a quadratic equation looks like this: \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable. This equation represents a parabola when graphed on a coordinate plane. The "quadratic" part comes from "quad", which means square, indicating the highest degree term is squared.

For example, in the equation \( 4t^2 - 8t - 1 = 0 \), the following holds:

• \( a \): The coefficient of \( t^2 \), which is 4
• \( b \): The coefficient of \( t \), which is -8
• \( c \): The constant term, which is -1

Understanding these components is crucial for solving quadratic equations as it helps set up the quadratic formula correctly.

Solving Quadratic Equations

To solve quadratic equations means finding the values of \( x \) (or in our original problem, \( t \)) that satisfy the equation. One popular method for solving is using the quadratic formula, which can solve any quadratic equation. The quadratic formula is:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula gives the solutions to the equation \( ax^2 + bx + c = 0 \). Here’s how you use it:

• Identify \( a \), \( b \), and \( c \): Recognize the coefficients from the equation.
• Substitute: Input these values into the quadratic formula.
• Calculate: Solve the equation using basic algebraic operations.

The solutions could be real numbers or complex numbers, depending on the discriminant \( b^2 - 4ac \). If the discriminant is positive, there are two real solutions; if zero, one real solution; and if negative, two complex solutions.

In this problem, by substituting \( a = 4 \), \( b = -8 \), and \( c = -1 \) into the formula, you can find \( t_1 = 1 + \frac{\sqrt{5}}{2} \) and \( t_2 = 1 - \frac{\sqrt{5}}{2} \) as the solutions.

Algebraic Expressions

Algebraic expressions are mathematical phrases that can include numbers, variables, and operations. In the quadratic context, expressions like \( 4t^2 - 8t - 1 \) combine constants \( 4 \), \(-8 \), \(-1 \) with the variable \( t \) to form an expression.

When solving equations, you often work with these expressions. Handling them involves simplifying and manipulating terms. For example, when using the quadratic formula, it's necessary to:

• Simplify Square Roots: Such as \( \sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5} \)
• Perform Basic Operations: Adding, subtracting, multiplying, or dividing terms, as seen in dividing \( 8 \pm 4\sqrt{5} \) by 8.
• Facilitate Factoring: While not directly used in the solution with the quadratic formula, factoring is another method to solve quadratics when applicable.

By understanding the structure and operations on algebraic expressions, you can confidently approach solving quadratic equations and other algebra problems.