Chapter 1: Problem 76
Find all real solutions of the quadratic equation. $$3+5 z+z^{2}=0$$
Short Answer
Expert verified
The real solutions are \( z = \frac{-5 + \sqrt{13}}{2} \) and \( z = \frac{-5 - \sqrt{13}}{2} \).
Step by step solution
01
Understanding the Problem
We are given a quadratic equation of the form \( z^2 + 5z + 3 = 0 \). Our task is to find all possible real solutions for \( z \).
02
Applying the Quadratic Formula
The quadratic formula is given by \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). In our equation, \( a = 1 \), \( b = 5 \), and \( c = 3 \). We will use these values to find the solutions.
03
Calculating the Discriminant
First, we calculate the discriminant \( \Delta \) which is \( b^2 - 4ac \). Here, \( b^2 = 25 \) and \( 4ac = 12 \), so \( \Delta = 25 - 12 = 13 \). Since \( \Delta > 0 \), there are two real solutions.
04
Finding the Roots
We now substitute \( a = 1 \), \( b = 5 \), and \( \sqrt{13} \) (from our discriminant) into the quadratic formula: \( z = \frac{-5 \pm \sqrt{13}}{2} \).
05
Calculating the Solutions
We have two solutions: \( z_1 = \frac{-5 + \sqrt{13}}{2} \) and \( z_2 = \frac{-5 - \sqrt{13}}{2} \). These are the real solutions of the quadratic equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
quadratic formula
The quadratic formula is a powerful tool for solving quadratic equations, which are equations that can be written in the form \( ax^2 + bx + c = 0 \). Here's how it works:
- Identify the coefficients \( a \), \( b \), and \( c \) in the equation. In our case, with the equation \( z^2 + 5z + 3 = 0 \), we have \( a = 1 \), \( b = 5 \), and \( c = 3 \).
- Plug these values into the quadratic formula: \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
discriminant
The discriminant is a crucial part of solving a quadratic equation using the quadratic formula. It is the term \( b^2 - 4ac \) found under the square root in the quadratic formula and helps determine the nature of the solutions.
- Calculate \( b^2 \), where \( b \) is the coefficient of the linear term. For our equation, \( b^2 = 25 \).
- Calculate \( 4ac \), which is four times the product of \( a \) and \( c \). Here, \( 4ac = 12 \).
- The discriminant is the difference \( b^2 - 4ac \). In this problem, it is \( 13 \).
- If \( \Delta > 0 \), there are two distinct real solutions.
- If \( \Delta = 0 \), there is one real, repeated solution.
- If \( \Delta < 0 \), there are no real solutions (but two complex ones).
real solutions
Real solutions are solutions of an equation that aren't imaginary; they are numbers that can be located on the number line. For our quadratic equation \( 3 + 5z + z^2 = 0 \), the calculation of the discriminant indicated two real solutions.Our quadratic formula yields the solutions:
- \( z_1 = \frac{-5 + \sqrt{13}}{2} \)
- \( z_2 = \frac{-5 - \sqrt{13}}{2} \)
- In physics, they can represent actual measurable quantities such as distance or time.
- In engineering, they might reflect practical dimensions or tolerances in designs.
- In statistics, real solutions may indicate observed or probable values.