Problem 83 - Algebra A Combined Function

Question

Evaluate \(\sqrt{b^{2}-4 a c}\) for each set of values. See Section \(10.3 .\) $$ a=2, b=4, c=-1 $$

Step-by-Step Solution

Answer

The value is \( 2\sqrt{6} \).

Step 1: Substitute the Values

Start by substituting the given values into the expression \( \sqrt{b^{2} - 4ac} \). Given values are \( a = 2 \), \( b = 4 \), and \( c = -1 \). Therefore, we substitute these into the expression to get \( \sqrt{4^2 - 4 \cdot 2 \cdot (-1)} \).

Step 2: Calculate the Square and Products

Calculate \( b^2 \), which is \( 4^2 = 16 \). Then calculate the product \( 4ac \), which is \( 4 \cdot 2 \cdot (-1) = -8 \).

Step 3: Simplify the Expression Inside the Square Root

Simplify the expression inside the square root: \( 16 - (-8) \). Simplifying the expression gives \( 16 + 8 = 24 \).

Step 4: Evaluate the Square Root

Now, calculate the square root of the simplified expression: \( \sqrt{24} \). Simplifying \( \sqrt{24} \) further gives \( 2\sqrt{6} \), since \( 24 = 4 \times 6 \) and \( \sqrt{4} = 2 \).

Key Concepts

Quadratic Formula

The quadratic formula is an essential concept in algebra that allows you to find the roots of any quadratic equation in the form \( ax^2 + bx + c = 0 \). The formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Here, \( a \), \( b \), and \( c \) are coefficients of the equation, and the term \( b^2 - 4ac \) is known as the "discriminant." This formula is universally applicable, and understanding how each part interacts is crucial for solving quadratic equations.

• The discriminant determines the nature of the roots (real or complex).
• The formula handles all cases (two, one, or zero real roots).

Using the formula, especially knowing that the discriminant can be analyzed alone to predict the nature of the roots, is a powerful algebraic tool.

Radical Expressions

Radical expressions involve roots such as square roots, cube roots, and others. Understanding them is crucial for algebra and calculus.

• The basic form is \( \sqrt[n]{x} \), where \( n \) is the root degree.
• When \( n = 2 \), it is a square root, which is the most commonly encountered radical.

Radicals can be both simplified and approximated. The square root of a perfect square, like \( \sqrt{16} = 4 \), is an example of simplification. Radical expressions are often involved in the quadratic formula, where the discriminant appears under a square root.

Simplifying Radicals

Simplifying radicals makes expressions cleaner and often easier to work with. It involves breaking down the number inside the radical into its prime factors and pairing factors for easier extraction outside the radical.

• For example, \( \sqrt{24} = 2\sqrt{6} \) because \( 24 = 4 \times 6 \) and \( \sqrt{4} = 2 \).
• Always look to extract square factors, since they simplify cleanly.

This simplification aids in solving equations and expressions efficiently. Ensure that you always reduce radicals to their simplest form, which often also simplifies subsequent calculations.

Algebraic Substitution

Algebraic substitution is a mathematical technique used to simplify calculations or solve equations. This concept often involves replacing variables with given numerical values or simpler expressions.

• In solving \( \sqrt{b^2 - 4ac} \), substitution was used to replace \( a = 2 \), \( b = 4 \), and \( c = -1 \).
• This results in a numerical expression that is easier to evaluate: \( \sqrt{16 - (-8)} = \sqrt{24} \).

Understanding algebraic substitution helps tackle complex expressions by breaking them down into manageable parts. It is a foundational skill in algebra, used broadly in calculus, physics, and engineering applications.