Problem 58 - Intermediate Algebra

Question

For each quadratic equation, first use the discriminant to determine whether the equation has two nonreal complex solutions, one real solution with a multiplicity of two, or two real solutions. Then solve the equation. $$ 8 x^{2}+18 x-5=0 $$

Step-by-Step Solution

Answer

Two distinct real solutions: \( x = \frac{1}{4} \) and \( x = -\frac{5}{2} \).

Step 1: Identify Coefficients

The given quadratic equation is \( 8x^2 + 18x - 5 = 0 \). Here, the coefficients are identified as follows: \( a = 8 \), \( b = 18 \), \( c = -5 \).

Step 2: Calculate the Discriminant

The discriminant of a quadratic equation \( ax^2 + bx + c = 0 \) is given by the formula \( D = b^2 - 4ac \). Substitute the identified coefficients into the formula: \( D = 18^2 - 4 \times 8 \times (-5) \). This simplifies to \( D = 324 + 160 = 484 \).

Step 3: Determine the Nature of Solutions

Since the discriminant \( D = 484 \) is positive and a perfect square, the equation has two distinct real solutions.

Step 4: Solve the Quadratic Equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{D}}{2a} \) to find the solutions. Substitute \( a = 8 \), \( b = 18 \), and \( D = 484 \) into the formula: \( x = \frac{-18 \pm \sqrt{484}}{16} \). This leads to \( x = \frac{-18 \pm 22}{16} \).

Step 5: Calculate the Solutions

There are two possibilities: \( x = \frac{-18 + 22}{16} = \frac{4}{16} = \frac{1}{4} \) and \( x = \frac{-18 - 22}{16} = \frac{-40}{16} = -\frac{5}{2} \). Thus, the solutions are \( x = \frac{1}{4} \) and \( x = -\frac{5}{2} \).

Key Concepts

Discriminant

In a quadratic equation of the form \( ax^2 + bx + c = 0 \), the **discriminant** helps us determine the nature of the solutions. It's calculated using the formula:

• \( D = b^2 - 4ac \)

The discriminant gives insight into whether solutions are real or complex:

• If \( D > 0 \), the equation has two distinct real solutions.
• If \( D = 0 \), there is exactly one real solution, sometimes called a repeated or double root.
• If \( D < 0 \), the solutions are complex (non-real).

In our problem, by substituting the coefficients, we calculated:

• \( D = 18^2 - 4 \times 8 \times (-5) = 484 \)

Since \( 484 \) is a positive number and a perfect square, the quadratic equation offers two different real solutions.

Real Solutions

Real solutions refer to the real-number answers of a quadratic equation. Quadratic equations can have:

• Two distinct real solutions when the discriminant \( D > 0 \).
• One real solution (a repeated root) when \( D = 0 \).

Real solutions are the points where the parabola (graph of the quadratic equation) intersects the x-axis. In our situation, where the discriminant \( D = 484 \), the equation offers two real numbers as solutions. This is because \( 484 \) is positive. The parabola crosses the x-axis at these two points, which are found using the quadratic formula. This gives the equation specific values where it holds true.

Quadratic Formula

The **quadratic formula** is a tool for finding the solutions of a quadratic equation, especially helpful when factoring is challenging. The formula is given by:

• \( x = \frac{-b \pm \sqrt{D}}{2a} \)

Here, \( b \) is the coefficient of \( x \), \( a \) is the leading coefficient, and \( D \) is the discriminant. The symbol \( \pm \) indicates that there are typically two solutions.To solve \( 8x^2 + 18x - 5 = 0 \) using the quadratic formula:

• Plug \( a = 8 \), \( b = 18 \), and \( D = 484 \) into the formula.
• Compute: \( x = \frac{-18 \pm \sqrt{484}}{16} \).
• Simplify to find \( x = \frac{-18 + 22}{16} = \frac{1}{4} \) or \( x = \frac{-18 - 22}{16} = -\frac{5}{2} \).

These steps provide the exact points where the parabola intersects the x-axis, representing the roots of the quadratic equation.