Problem 19 - Algebra for College Students

Question

Solve each quadratic equation using the method that seems most appropriate to you. $$4 t^{2}+4 t-1=0$$

Step-by-Step Solution

Answer

Solutions: \( t = \frac{-1 + \sqrt{2}}{2} \) and \( t = \frac{-1 - \sqrt{2}}{2} \).

Step 1: Identify the quadratic equation

The given quadratic equation is in the form \( ax^2 + bx + c = 0 \), where \( a = 4 \), \( b = 4 \), and \( c = -1 \).

Step 2: Apply the quadratic formula

The quadratic formula is \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute \( a = 4 \), \( b = 4 \), and \( c = -1 \) into the formula.

Step 3: Calculate the discriminant

First, calculate the discriminant \( b^2 - 4ac \): \ \( b^2 = 4^2 = 16 \) \ \( 4ac = 4 \times 4 \times (-1) = -16 \) \ Thus, the discriminant is \( 16 - (-16) = 16 + 16 = 32 \).

Step 4: Solve for t

Use the result from the discriminant to solve for \( t \): \ \( t = \frac{-4 \pm \sqrt{32}}{2 \times 4} \) \ \( t = \frac{-4 \pm \sqrt{32}}{8} \).

Step 5: Simplify the square root and the expression

Simplify \( \sqrt{32} \) to \( 4\sqrt{2} \), \ then the expression becomes: \ \( t = \frac{-4 \pm 4\sqrt{2}}{8} \). \ Factor out 4 from the numerator: \( t = \frac{4(-1 \pm \sqrt{2})}{8} = \frac{-1 \pm \sqrt{2}}{2} \).

Step 6: Present the solution

Thus, the solutions to the quadratic equation are \( t = \frac{-1 + \sqrt{2}}{2} \) and \( t = \frac{-1 - \sqrt{2}}{2} \).

Key Concepts

Quadratic Formula

To solve quadratic equations, one reliable method is the Quadratic Formula. It's especially helpful when factoring is not straightforward. The formula is given by:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]It applies to any quadratic equation in the standard form \( ax^2 + bx + c = 0 \). Here:

• \( a \) is the coefficient of \( t^2 \)
• \( b \) is the coefficient of \( t \)
• \( c \) is the constant term

To use the formula, identify these coefficients in your quadratic equation. Substitute them into the formula, and simplify to find the roots. This approach works for all quadratic equations, whether they have real or complex solutions.

Discriminant in Quadratics

In any quadratic equation, the discriminant, denoted as \( b^2 - 4ac \), plays a crucial role. It tells us about the nature and number of the solutions:

• If the discriminant is positive, we have two different real roots.
• If it is zero, there is exactly one real root (also known as a double root).
• If the discriminant is negative, the roots are complex and occur as a pair of conjugates.

Calculating the discriminant is an essential part of solving quadratic equations with the quadratic formula. It helps you understand what kind of solutions to expect:- For our equation, \( 4t^{2}+4t-1=0 \), we calculated the discriminant as 32.- As this is positive, it indicates two distinct real solutions.Thus, understanding and computing the discriminant can direct your method for solving quadratic equations.

Solving Quadratics Step-by-Step

Solving quadratic equations comprehensively involves several steps, especially when using the quadratic formula. Let's break it down:1. **Identify the Equation**: Begin by ensuring your given equation is in the form \( ax^2 + bx + c = 0 \). - For example, in \( 4t^2 + 4t - 1 = 0 \), \( a = 4 \), \( b = 4 \), and \( c = -1 \).2. **Calculate the Discriminant**: Use \( b^2 - 4ac \) to determine solution type. - Here, \( 16 + 16 = 32 \) is positive, indicating two real roots.3. **Plug into the Quadratic Formula**: Insert \( a \), \( b \), and \( c \) into \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). - You'll initially get \( t = \frac{-4 \pm \sqrt{32}}{8} \).4. **Simplify**: Break down \( \sqrt{32} \) to \( 4\sqrt{2} \) and simplify further. - Simplifies to \( t = \frac{-1 \pm \sqrt{2}}{2} \), providing the roots \( \frac{-1 + \sqrt{2}}{2} \) and \( \frac{-1 - \sqrt{2}}{2} \).These steps enable you to methodically solve any quadratic equation, ensuring both accuracy and understanding of the solutions obtained.