Problem 7 - Algebra for College Students

Question

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. $$15 x^{2}+17 x-4=0$$

Step-by-Step Solution

Answer

The equation has two real solutions: \( x = \frac{1}{5} \) and \( x = -\frac{4}{3} \).

Step 1: Identify the Coefficients

First, identify the coefficients in the quadratic equation \( 15x^2 + 17x - 4 = 0 \). The standard form is \( ax^2 + bx + c = 0 \). Here, \( a = 15 \), \( b = 17 \), and \( c = -4 \).

Step 2: Calculate the Discriminant

The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by the formula \( D = b^2 - 4ac \). Substitute the values of \( a \), \( b \), and \( c \) into the formula to find \( D \). \[ D = 17^2 - 4 \cdot 15 \cdot (-4) \]\[ D = 289 + 240 \]\[ D = 529 \]

Step 3: Determine Nature of Solutions

Interpret the discriminant value. Since \( D = 529 \) and \( D > 0 \), the quadratic has two distinct real solutions.

Step 4: Use the Quadratic Formula

To find the solutions, use the quadratic formula: \( x = \frac{{-b \pm \sqrt{D}}}{{2a}} \). With \( D = 529 \), substitute into the formula.\[ x = \frac{{-17 \pm \sqrt{529}}}{{2 \cdot 15}} \]\[ x = \frac{{-17 \pm 23}}{{30}} \]

Step 5: Calculate Both Solutions

Solve for the two solutions from the quadratic formula.1. For the positive root: \[ x = \frac{{-17 + 23}}{{30}} = \frac{6}{30} = \frac{1}{5} \]2. For the negative root: \[ x = \frac{{-17 - 23}}{{30}} = \frac{-40}{30} = -\frac{4}{3} \]

Step 6: Verify the Solutions

Confirm the solutions satisfy the original equation:- Substituting \( x = \frac{1}{5} \) yields zero when plugged into the equation.- Similarly, substituting \( x = -\frac{4}{3} \) also results in zero, confirming both solutions are correct.

Key Concepts

Discriminant

The discriminant is a key feature in quadratic equations used to determine the nature of their roots. It is calculated using the formula \( D = b^2 - 4ac \), where \( a \), \( b \), and \( c \) are coefficients from the standard form of a quadratic equation \( ax^2 + bx + c = 0 \). The discriminant tells us whether the solutions of the quadratic are real or complex:

• If \( D > 0 \), the equation has two distinct real solutions.
• If \( D = 0 \), there is exactly one real solution, also known as a repeated or double root.
• If \( D < 0 \), the solutions are complex and appear in conjugate pairs.

In our example, the discriminant calculated is 529, a positive number, indicating that there are two distinct real solutions. This gives us an insight into the solution before we even start with actual calculations.

Quadratic Formula

The quadratic formula is a reliable method for finding solutions to any quadratic equation. This formula is written as \( x = \frac{{-b \pm \sqrt{D}}}{{2a}} \), where \( D \) is the discriminant. The formula provides two solutions for \( x \), which are:

• The plus sign (\(+\)) in front of the square root gives the first solution.
• The minus sign (\(-\)) provides the second solution.

When applying the quadratic formula, it's essential to ensure the values of \( a \), \( b \), and \( D \) are substituted correctly. In the example exercise, \( a = 15 \), \( b = 17 \), and the discriminant \( D = 529 \). Substituting into the formula, you calculate:
For the first solution: \( x = \frac{{-17 + 23}}{{30}} = \frac{1}{5} \)
For the second solution: \( x = \frac{{-17 - 23}}{{30}} = -\frac{4}{3} \)
These solutions represent the points where the parabola described by the quadratic equation intersects the x-axis.

Real and Complex Solutions

The nature of the solutions of a quadratic equation is dictated by its discriminant. In mathematics, understanding whether a quadratic will produce real or complex solutions is crucial for determining subsequent steps in problem-solving.
- **Real Solutions:** - Two distinct real solutions occur when the discriminant \( D > 0 \). This indicates that the graph of the quadratic crosses the x-axis at two different points. - One real solution happens when \( D = 0 \), meaning the graph touches the x-axis at a single point (vertex). This is called a vertex or repeated solution.- **Complex Solutions:** - When \( D < 0 \), the quadratic has two complex solutions. These do not intersect the x-axis at real number points instead, they form a complex conjugate pair. Complex solutions are important in scenarios involving oscillations and waves.In our specific problem, since \( D = 529 \) is larger than zero, we have two distinct real solutions: \( x = \frac{1}{5} \) and \( x = -\frac{4}{3} \), confirming the nature as predicted by the discriminant.