91Ó°ÊÓ

The quadratic formula is a versatile tool for solving quadratic equations. A quadratic equation is any equation of the form \ ax^2 + bx + c = 0 \. When we reference the quadratic formula, we mean the equation: \ y = \frac{ -b \pm \sqrt{b^2 - 4ac} }{2a} \. This formula can be used to solve any quadratic equation by substituting the respective coefficients \( a \), \( b \), and \( c \) from your quadratic equation.
In our exercise: the quadratic equation was \( 2y^2 - 2y + 1 = 0 \). Here, \( a \) is 2, \( b \) is -2, and \( c \) is 1. Substitute these values into the quadratic formula to find \( y \). This helps to find the roots or solutions of the quadratic equation.
An important thing to note is the discriminant: the part of the formula underneath the square root sign, \( b^2 - 4ac \). The discriminant tells us about the nature of the two solutions:

• If \( b^2 - 4ac > 0 \), there are two distinct real solutions.
• If \( b^2 - 4ac = 0 \), there is one real solution.
• If \( b^2 - 4ac < 0 \), the solutions are complex or imaginary.

The substitution method is a handy algebraic technique used to simplify complex equations. In our case, we started with a rational equation: \ \frac{1}{(x+1)^2} - \frac{2}{x+1} + 2 = 0 \. To make it easier to work with, we can introduce a new variable \( y \) where \( y = x + 1 \). This transformation simplifies the equation to a more straightforward quadratic form: \( \frac{1}{y^2} - \frac{2}{y} + 2 = 0 \).
Substitution helps in reducing the equation's complexity and makes it easier to solve.

• Step 1: Define the new variable based on the problem. Here, it was \( y = x + 1 \).
• Step 2: Replace the old variable \( x \). This simplifies the structure of the equation.
• Step 3: Solve the new equation using appropriate methods (in this case, the quadratic formula).
• Step 4: Substitute back the original variable to find the final solution.

This systematic approach is particularly helpful in transforming non-linear equations into solvable linear or quadratic forms.

Complex numbers are an extension of the real numbers, and they introduce a new unit, \( i \), which is the imaginary unit. The imaginary unit is defined as \( i = \sqrt{-1} \). Therefore, complex numbers have a real part and an imaginary part of the form \( a + bi \), where \( a \) and \( b \) are real numbers.

• The real part is simply the portion without the imaginary unit, \( a \).
• The imaginary part is the portion with the imaginary unit, \( bi \).

In our exercise, we found the imaginary solutions as \( y = \frac{1 \pm i}{2} \). Understanding complex numbers opens the door to solving equations that do not have real solutions. They are crucial for higher-level mathematics as well as engineering, physics, and other sciences.

When solving quadratic equations, especially when the discriminant \( b^2 - 4ac \) is negative, the solutions turn out to be imaginary. Imaginary solutions are those that involve the imaginary unit \( i \), which itself is equal to the square root of \(-1\).
In our exercise, we derived \( y \) values that included imaginary numbers. For instance, when we calculated \( y = \frac{ 2 \pm \sqrt{ -4 }}{4} \), we got imaginary solutions because \( \sqrt{-4} \ = 2i \):

• \( y = \frac{1 + i}{2} \)
• \( y = \frac{1 - i}{2} \)

Upon back-substitution to find \( x \), we found the imaginary solutions for the original equation. These imaginary solutions are vital in fields such as electrical engineering, control theory, and even quantum physics. They illustrate how complex some equations' solutions can be beyond just real numbers.