91Ó°ÊÓ

The quadratic formula helps us solve equations of the form \text{ax}^2 + \text{bx} + \text{c} = 0. It gives us the solutions, or roots, of the quadratic equation. The formula is given as: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In this formula, \(a\), \(b\), and \(c\) are constants from the quadratic equation.
The term under the square root, \(b^2 - 4ac\), is called the discriminant. The discriminant tells us the nature of the roots:

• If \(b^2 - 4ac > 0\), we get two distinct real roots.
• If \(b^2 - 4ac = 0\), we get one real root (a repeated root).
• If \(b^2 - 4ac < 0\), the roots are nonreal complex (involving imaginary numbers).

In our exercise, we used this formula to solve \( 5y^2 - 12y + 7 = 0 \). By substituting \(a = 5\), \(b = -12\), and \(c = 7\), we calculated the discriminant to be 4, indicating two real solutions for \(y\). Then, by evaluating the quadratic formula, we found \(y = 1.4\) and \(y = 1\).

Nonreal complex solutions occur in quadratic equations when the discriminant (\(b^2 - 4ac\)) is negative. This means that the equation does not intersect the x-axis and has no real number solutions. Instead, the roots are complex numbers and include the imaginary unit \(i\), where \(i = \sqrt{-1}\). The solutions to such a quadratic equation can be written as: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] When the term under the square root is negative, it introduces the imaginary unit \(i\). For example, for a quadratic equation where the discriminant \(b^2 - 4ac = -4\), the roots would be: \[ y = \frac{-b \pm \sqrt{-4}}{2a} = \frac{-b \pm 2i}{2a} \] This results in complex conjugate solutions such as \(y = -b/(2a) + i/a\) and \(y = -b/(2a) - i/a\).
In our exercise, the discriminant was positive, so we did not encounter nonreal complex solutions. However, understanding this concept is vital for solving any quadratic equation.

A system of equations consists of multiple equations that we solve simultaneously. The solutions are the values of the variables that make all the equations true at the same time. Systems can be linear or nonlinear. In our exercise, we dealt with a nonlinear system: \[ \begin{aligned} 3x^2 + 2y^2 & = 5 \ x - y & = -2 \end{aligned} \] Here, one equation is quadratic, making the system nonlinear.
To solve this system:

• Solve the linear equation for one variable. We solved \(x - y = -2\) for \(x = y - 2\).
• Substitute the expression from the linear equation into the nonlinear equation. We substituted \(x = y - 2\) into \(3(y - 2)^2 + 2y^2 = 5\).
• Simplify and solve the resulting quadratic equation. This resulted in \(5y^2 - 12y + 7 = 0\).
• Find the values for the other variable. Using \(y = 1.4\) and \(y = 1\), we found corresponding \(x\) values by substituting back into the equation \(x = y - 2\).

This approach ensures that all parts of the system are satisfied, providing us with valid solutions.