Problem 25 - Algebra A Combined Function

Question

Use the quadratic formula to solve each equation. These equations have real solutions and complex, but not real, solutions. $$ (x+5)(x-1)=2 $$

Step-by-Step Solution

Answer

The solutions are \(x = -2 \pm \sqrt{11}\).

Step 1: Expand the Equation

Expand the left-hand side of the equation (x+5)(x-1) using the distributive property: \((x+5)(x-1) = x^2 - x + 5x - 5 = x^2 + 4x - 5\).

Step 2: Set Equation to Standard Quadratic Form

Set the equation equal to zero by moving the 2 from the right-hand side to the left:\(x^2 + 4x - 5 - 2 = 0\), which simplifies to \(x^2 + 4x - 7 = 0\).

Step 3: Identify Coefficients for the Quadratic Formula

Identify the coefficients in the quadratic equation \(ax^2 + bx + c = 0\). Here, \(a = 1\), \(b = 4\), and \(c = -7\).

Step 4: Apply the Quadratic Formula

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with the identified coefficients:\(x = \frac{-(4) \pm \sqrt{(4)^2 - 4 \cdot 1 \cdot (-7)}}{2 \cdot 1}\).

Step 5: Calculate the Discriminant

Calculate the discriminant \(b^2 - 4ac\):\((4)^2 - 4 \cdot 1 \cdot (-7) = 16 + 28 = 44\).

Step 6: Compute the Roots Using the Quadratic Formula

Substitute the discriminant back into the quadratic formula:\(x = \frac{-4 \pm \sqrt{44}}{2}\). Simplify \(\sqrt{44} = \sqrt{4 \times 11} = 2\sqrt{11}\), so:\(x = \frac{-4 \pm 2\sqrt{11}}{2}\).

Step 7: Simplify the Expression

Simplify the expression to find the roots:\[x = \frac{-4}{2} \pm \frac{2\sqrt{11}}{2} = -2 \pm \sqrt{11}\].

Key Concepts

Quadratic Equations

Quadratic equations are a fundamental concept in algebra that describe equations of the form \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) are constants with \(a eq 0\). A quadratic equation can be identified by its distinctive highest-degree term, \(x^2\). To solve these equations, several methods can be employed including:

• Factoring (if applicable)
• Completing the square
• Using the quadratic formula

The quadratic formula is famously given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula is used universally for solving any quadratic equation, especially when factoring is difficult or impossible. To effectively use the quadratic formula, it is crucial to identify the coefficients \(a\), \(b\), and \(c\) from the equation's standard form and substitute these values into the formula.

Complex Solutions

While many quadratic equations have real-number solutions, some result in complex solutions. This typically occurs when the discriminant (the expression inside the square root of the quadratic formula) is negative. Complex numbers are in the form \(a + bi\) where \(i\) is the imaginary unit and \(i^2 = -1\).
Quadratics with complex solutions will have two conjugate solutions, such as \(x = a + bi\) and \(x = a - bi\).
Even if solutions are complex, the quadratic equation actually has definite roots:

• The part "\(a\)" gives its real component.
• The part "\(bi\)" provides its imaginary component making it an important concept in algebra.

Mastering the concept of complex solutions is essential in mathematics, particularly in higher-level calculations involving polynomials and systems that don’t have real solutions.

Discriminant

The discriminant is a crucial part of the quadratic formula and plays a significant role in determining the nature of the equation's roots. It is represented by the expression:\[b^2 - 4ac\]Depending on the value of the discriminant, the type of solutions for the quadratic equation can be predicted:

• If the discriminant is positive, the quadratic equation has two distinct real solutions.
• If it is zero, there is one real solution, which corresponds to a double root.
• If the discriminant is negative, the solutions are complex or imaginary, and not real.

Understanding the discriminant helps us anticipate the solution's form without fully solving the equation and is integral in exploring the properties of quadratic equations.