Chapter 11: Problem 40
Use the quadratic formula to solve each equation. These equations have real solutions and complex, but not real, solutions. See Examples 1 through 4. $$ \left(p-\frac{1}{2}\right)^{2}=\frac{p}{2} $$
Short Answer
Expert verified
The solutions are \( p = \frac{3 + \sqrt{5}}{4} \) and \( p = \frac{3 - \sqrt{5}}{4} \).
Step by step solution
01
Expand the Left Side
Start by expanding the left side of the equation: \( \left(p - \frac{1}{2}\right)^{2} = \frac{p}{2} \). This can be rewritten as \( \left(p - \frac{1}{2}\right)(p - \frac{1}{2}) = p^2 - p + \frac{1}{4} \).
02
Rearrange the Equation
Bring \( \frac{p}{2} \) from the right side to the left side of the equation: \( p^2 - p + \frac{1}{4} - \frac{p}{2} = 0 \). Simplify it to: \( p^2 - \frac{3}{2}p + \frac{1}{4} = 0 \).
03
Use the Quadratic Formula
The quadratic formula is: \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -\frac{3}{2} \), and \( c = \frac{1}{4} \). First, calculate the discriminant: \( b^2 - 4ac = \left(-\frac{3}{2}\right)^2 - 4(1)\left(\frac{1}{4}\right) = \frac{9}{4} - 1 = \frac{5}{4} \).
04
Calculate the Roots
Substitute the values into the quadratic formula to find the roots: \(p = \frac{-(-\frac{3}{2}) \pm \sqrt{\frac{5}{4}}}{2(1)}\). Simplify to get \( p = \frac{3 \pm \sqrt{5}}{4} \). The solutions are \( p = \frac{3 + \sqrt{5}}{4} \) and \( p = \frac{3 - \sqrt{5}}{4} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Complex Solutions
In algebra, the concept of complex solutions arises when solving quadratic equations. A quadratic equation can sometimes have solutions that are not just real numbers but complex numbers. A complex number consists of a real part and an imaginary part. Imaginary numbers are multiples of the imaginary unit, denoted as \(i\), where \(i^2 = -1\). When we solve a quadratic equation using the quadratic formula,
- we sometimes encounter a negative number under the square root within the formula.
- This results in complex solutions, as the square root of a negative number is imaginary.
Discriminant
The discriminant is a specific expression derived from the quadratic formula and plays a crucial role in determining the nature of the solutions of a quadratic equation. It is represented by the expression \(b^2 - 4ac\).
- If the discriminant is positive, a quadratic equation has two distinct real solutions.
- If it is zero, there is exactly one real solution (or a repeated real solution).
- If the discriminant is negative, the quadratic equation has complex solutions.
Quadratic Equation
A quadratic equation is a second-degree polynomial equation in a single variable. Its standard form is expressed as \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\).
- The term \(ax^2\) is the quadratic term.
- \(bx\) is the linear term, and \(c\) is the constant term.
Real Solutions
Real solutions of a quadratic equation are the values of the variable that satisfy the equation and are real numbers. When the discriminant \(b^2 - 4ac\) is greater than or equal to zero, the quadratic equation has real solutions.
- A positive discriminant results in two distinct real solutions.
- A zero discriminant results in a double real solution, often called a repeated root.