Chapter 11: Problem 37
Use the quadratic formula to solve each equation. These equations have real solutions and complex, but not real, solutions. $$ \frac{1}{2} y^{2}=y-\frac{1}{2} $$
Short Answer
Expert verified
The solution is \( y = 1 \), a repeated real root.
Step by step solution
01
Rewrite the equation in standard form
The given equation is \( \frac{1}{2} y^{2} = y - \frac{1}{2} \). Before applying the quadratic formula, it is important to rearrange it to the standard form \( ay^{2} + by + c = 0 \). Multiply every term by 2 to get \( y^{2} = 2y - 1 \). Move all terms to one side to get the equation \( y^{2} - 2y + 1 = 0 \).
02
Identify coefficients a, b, and c
For the equation \( y^{2} - 2y + 1 = 0 \), identify coefficients: \( a = 1 \), \( b = -2 \), and \( c = 1 \).
03
Apply the quadratic formula
The quadratic formula is \( y = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \). Substitute \( a = 1 \), \( b = -2 \), and \( c = 1 \) into the formula.
04
Calculate the discriminant
First, calculate the discriminant \( b^{2} - 4ac \). Here \( b^2 = (-2)^2 = 4 \) and \( 4ac = 4(1)(1) = 4 \). So, \( b^{2} - 4ac = 4 - 4 = 0 \).
05
Solve for y using the quadratic formula
Since the discriminant is 0, this means there is exactly one real solution (a repeated root). Substitute the discriminant and other values into the quadratic formula: \( y = \frac{-(-2) \pm \sqrt{0}}{2 \times 1} = \frac{2}{2} = 1 \). So, the solution is \( y = 1 \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equations
Quadratic equations are a fundamental part of algebra and can be expressed in the standard form: \[ ax^2 + bx + c = 0 \]. Here, \( a \), \( b \), and \( c \) represent constants, where \( a eq 0 \). The variable \( x \) is what we're solving for, known as the unknown or variable. Every quadratic equation can graphically be represented as a parabola, a U-shaped curve, on a coordinate plane.
- If \( a > 0 \), the parabola opens upwards.
- If \( a < 0 \), the parabola opens downwards.
Discriminant
The discriminant is a key component in the quadratic formula, denoted as \( b^2 - 4ac \). It provides crucial information about the nature and number of the solutions of a quadratic equation. By examining the discriminant:
- If \( b^2 - 4ac > 0 \), the equation has two distinct real solutions.
- If \( b^2 - 4ac = 0 \), there is one real solution, often referred to as a repeated or double root.
- If \( b^2 - 4ac < 0 \), the solutions are non-real (complex) and occur as complex conjugates.
Real Solutions
Real solutions of quadratic equations are those that can be plotted on the real number line. When solving a quadratic equation using the quadratic formula, real solutions occur when the discriminant \( b^2 - 4ac \) is zero or positive. In our example, the discriminant was found to be 0, which implies a real solution. The math behind it shows that the quadratic equation reduces to having a single solution, what we call a repeated root. This means that the point on the parabola just touches the x-axis at this one value.For the example equation, the real solution is \( y = 1 \), and this shows that the parabola touches the x-axis at this single point, emphasizing the concept of repeated roots.
Complex Solutions
Complex solutions arise when the discriminant \( b^2 - 4ac \) of a quadratic equation is negative. In such cases, the solutions include imaginary numbers, indicated because the square root of a negative number involves the imaginary unit \( i \). Mathematically, when you solve such equations, the solutions take the form:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( \sqrt{b^2 - 4ac} \) becomes an imaginary number because of the negative discriminant.These solutions, often referred to as complex conjugates, come in pairs like \( a + bi \) and \( a - bi \). While they don't correspond to any point on the real number line, they provide essential information about the properties of a quadratic graph. For instance, they can denote how a parabola behaves in relation to the x-axis, such as not intersecting it at any point. However, in our given example, complex solutions don't occur as the discriminant was zero, resulting in real solutions.