91Ó°ÊÓ

Quadratic equations are fundamental in algebra and can be identified by their general form: \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. An essential characteristic of quadratic equations is the squaring of the variable \(x\), which results in a parabolic graph when plotted. These parabolas can open upwards or downwards depending on the coefficient \(a\).

• If \(a > 0\), the parabola opens upwards.
• If \(a < 0\), it opens downwards.

Quadratic equations can have two real solutions, one real solution, or no real solutions, determined by the discriminant \(b^2 - 4ac\). Solving these equations often involves methods like factoring, completing the square, or using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). In the exercise, after substitution, the circle's equation is simplified into a quadratic equation \(5x^2 - 10x = 0\). We solve it by factoring to find what values of \(x\) satisfy the equation.

Linear equations represent straight lines on a graph and are usually in the form \(ax + by + c = 0\). A simple and common form to use is \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. These equations are first-degree, meaning the highest power of the variable is 1. They result in linear graphs which create straight lines.

In the system of equations given, the linear equation is \(2x - y - 3 = 0\). We rearrange it to \(y = 2x - 3\) to better utilize it within the substitution method. Understanding how to manipulate these equations is crucial, as they form the basis for finding intersections with other equations, such as quadratic equations or circles, in systems of equations.

The substitution method is a technique used to solve systems of equations. It involves expressing one variable in terms of another and substituting it into another equation. This helps reduce the system to a single equation with one variable, making it easier to solve.

Here's how you can effectively use the substitution method:

• First, solve one of the equations for one variable in terms of the other.
• Substitute this expression into the other equation, replacing the variable.
• Solve the resulting equation for the single variable.
• Finally, substitute back to find the other variable's value.

In our exercise, after expressing \(y\) as \(2x - 3\) from the linear equation, we substituted \(y\) into the circle equation. Then, we simplified and solved the resulting quadratic equation to find the values of \(x\). This method simplifies complex systems by breaking them into easier steps.

A circle's equation in algebra is often given as \((x-h)^2 + (y-k)^2 = r^2\). Here, \((h,k)\) represents the circle's center, and \(r\) its radius. This equation highlights the set of all points \( (x, y) \) that are equidistant from the center of the circle. When expanded or modified, it may appear differently but can always be rearranged to this typical form.

In the given exercise, the initial circle equation is \((x-1)^2 + (y+1)^2 = 5\). Upon simplification, it becomes a form we can use to solve the system of equations jointly with the linear equation. Understanding circle equations allows us to determine intersections, such as where a line intersects a circle, which involves using both linear and quadratic solving techniques.