Chapter 6: Problem 7
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. $$15 x^{2}+17 x-4=0$$
Short Answer
Expert verified
The equation has two real solutions: \( x = \frac{1}{5} \) and \( x = -\frac{4}{3} \).
Step by step solution
01
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 \( c = -4 \).
02
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 \( D \). \[ D = 17^2 - 4 \cdot 15 \cdot (-4) \]\[ D = 289 + 240 \]\[ D = 529 \]
03
Determine Nature of Solutions
Interpret the discriminant value. Since \( D = 529 \) and \( D > 0 \), the quadratic has two distinct real solutions.
04
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.\[ x = \frac{{-17 \pm \sqrt{529}}}{{2 \cdot 15}} \]\[ x = \frac{{-17 \pm 23}}{{30}} \]
05
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: \[ x = \frac{{-17 - 23}}{{30}} = \frac{-40}{30} = -\frac{4}{3} \]
06
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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:
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.
- The plus sign (\(+\)) in front of the square root gives the first solution.
- The minus sign (\(-\)) provides the second solution.
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.
- **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.