Problem 37 - Algebra and Trigonometry

Question

Solve each system by the method of your choice. $$ \left\\{\begin{array}{l} {x^{2}+(y-2)^{2}=4} \\ {x^{2}-2 y=0} \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system are the points (0,0), \((\sqrt{8/3}, 4/3)\), \((-\sqrt{8/3}, 4/3)\)

Step 1: Substitution

Substitute the value of \(x^{2}\) from the second equation into the first one. The second equation can be written as \(x^{2}=2y\). Substituting into the first equation, we get: \[(2y+(y-2)^{2}=4)\]

Step 2: Solve the Equation

Simplify the equation obtained above and make it as a quadratic equation. The equation simplifies to \[(3y^{2}-4y = 0)\] This is a quadratic equation in variable \(y\) and can be solved by factorizing. Factorizing yields: \[(y(3y-4) = 0)\]

Step 3: Find the Roots

Set the factors equal to zero and solve for \(y\). This gives the roots for the equation as: \[(y= 0, y=4/3)\]

Step 4: Substitute back

Substitute the obtained y values back into the second equation to determine the corresponding x values: Plug \(y = 0\) into \(x^{2} = 2y\), we get \(x = 0\). Plug \(y = 4/3\) into \(x^{2} = 2y\), we get \(x = \sqrt{8/3}, x = -\sqrt{8/3}\).

Step 5: Solutions of the System

The system then has three solutions, corresponding to the points where the circle intersects the parabola: \[(x=0, y=0), (x=\sqrt{8/3}, y=4/3), (x=-\sqrt{8/3}, y=4/3).\]

Key Concepts

Substitution Method

The substitution method is a powerful tool for solving systems of equations. It involves solving one of the equations for one variable, then substituting that expression into the other equation. This allows you to reduce the number of variables. In our example, we began with two equations involving variables \(x\) and \(y\):

• \(x^2 + (y-2)^2 = 4\)
• \(x^2 - 2y = 0\)

To use substitution, we first solve the second equation for \(x^2\), which gives \(x^2 = 2y\). This is then substituted into the first equation, replacing \(x^2\), to form a new equation in terms of \(y\) only. This method effectively reduces the problem to a simpler one-variable equation, making it easier to solve.

Quadratic Equations

Quadratic equations are polynomial equations of degree two, typically in the form \(ax^2 + bx + c = 0\). These equations can have two solutions due to their parabolic graph shape. In the exercise, substituting \(x^2 = 2y\) into \((x^2 + (y-2)^2 = 4)\), simplifies the problem to a quadratic equation in \(y\):

• \(3y^2 - 4y = 0\)

This is a standard form of a quadratic equation, easily recognizable and solvable by various methods such as factoring, completing the square, or applying the quadratic formula. Here, we choose to factor the equation, because it is one of the most straightforward methods and often requires simple algebraic manipulation.

Factoring Quadratics

Factoring quadratics is one of the simplest and most intuitive methods to solve quadratic equations. It involves breaking down a quadratic equation into a product of two binomials. The key step in the example was rearranging \((3y^2 - 4y = 0)\) so that it can be expressed as a product:

• \(y(3y - 4) = 0\)

By setting each factor equal to zero, \(y = 0\) and \(3y - 4 = 0\), we can easily determine the solutions for \(y\). This method works particularly well for equations that can be factored into integers or simple expressions, as it quickly reveals the roots or solutions of the equation.

Intersection of Curves

The concept of intersection of curves refers to finding the common points where two curves meet on a graph. In the context of solving a system of equations, each equation represents a curve. The solutions to the system are the points at which these curves intersect. In this exercise, one equation represents a circle and the other a parabola. These distinct curves intersect at:

• \((x=0, y=0)\)
• \((x=\sqrt{8/3}, y=4/3)\)
• \((x=-\sqrt{8/3}, y=4/3)\)

These points signify the solutions to the system, indicating where the algebraic expressions satisfy both equations simultaneously. Understanding this concept helps visualize how two or more equations relate in their graphical form and how solutions can be comprehended as tangible points of intersection in space.