Problem 22 - Algebra and Trigonometry

Question

Two equations and their graphs are given. Find the intersection point(s) of the graphs by solving the system. \(\left\\{\begin{aligned} x^{2}+y^{2} &=4 x \\ x &=y^{2} \end{aligned}\right.\)

Step-by-Step Solution

Answer

The intersection points are \((0, 0)\), \((3, \sqrt{3})\), and \((3, -\sqrt{3})\).

Step 1: Understand the Equations

We are given two equations: \(x^2 + y^2 = 4x\) and \(x = y^2\). Our goal is to find the intersection point(s) of their graphs by solving the system of equations.

Step 2: Solve for \(x\) in terms of \(y\) using the second equation

From the second equation, we already have \(x = y^2\). This will help us substitute into the first equation.

Step 3: Substitute \(x = y^2\) into the first equation

Substitute \(x = y^2\) into the first equation \(x^2 + y^2 = 4x\). This gives \((y^2)^2 + y^2 = 4(y^2)\). Simplifying, we get \(y^4 + y^2 = 4y^2\).

Step 4: Simplify the Equation

Rearrange the equation from Step 3: \(y^4 + y^2 = 4y^2\). Subtract \(4y^2\) from both sides to obtain \(y^4 - 3y^2 = 0\).

Step 5: Factor the Equation

The equation \(y^4 - 3y^2 = 0\) can be factored as \(y^2(y^2 - 3) = 0\). Thus, \(y^2 = 0\) or \(y^2 = 3\).

Step 6: Solve for \(y\) Values

From \(y^2 = 0\), we get \(y = 0\). From \(y^2 = 3\), we get \(y = \sqrt{3}\) and \(y = -\sqrt{3}\).

Step 7: Solve for Corresponding \(x\) Values

Using the equation \(x = y^2\):- For \(y = 0\), \(x = 0^2 = 0\).- For \(y = \sqrt{3}\), \(x = (\sqrt{3})^2 = 3\).- For \(y = -\sqrt{3}\), \(x = (-\sqrt{3})^2 = 3\).

Step 8: Determine the Intersection Points

The solutions \((x, y)\) are the intersection points of the graphs. The points are \((0, 0)\), \((3, \sqrt{3})\), and \((3, -\sqrt{3})\).

Key Concepts

Intersection Points in a System of Equations

In a system of equations involving two or more equations, intersection points are values of the variables that satisfy all equations simultaneously. These points represent where the curves or lines, described by the equations, meet on a graph.
To find these points, you need to solve the system of equations. This means finding values for the variables that make all equations true at the same time. In graphical terms, you're looking for the points where both graphs overlap or intersect.
When dealing with nonlinear equations, like in the exercise where one of the equations is quadratic, you might end up with multiple intersection points. Each of these solutions corresponds to a point on the graph where the two curves share (or intersect) the same coordinates in terms of both x and y. This means that both equations have the same solution at these specific points.

Understanding Quadratic Equations

Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). These equations form parabolic shapes when graphed.
The foundational property of quadratic equations is that they can have up to two real solutions, depending on the discriminant \( b^2 - 4ac \):

• If the discriminant is positive, there are two distinct real solutions.
• If the discriminant is zero, there is exactly one real solution.
• If the discriminant is negative, there are no real solutions.

In the context of a system of equations, as with the exercise given, solving for values of \(x\) when dealing with quadratic curves can yield multiple points of intersection, illustrating the solutions to the system of equations in graphical form.

Graphical Solutions of Equations

Graphical solutions involve visually representing equations on a coordinate plane and finding where they intersect. This method is especially useful for solving systems of equations, as it provides a clear visual depiction of the solutions.
For example, when a quadratic equation is involved, its graph is a parabola. In our exercise, one equation is a parabola, and the other, a quadratic in terms of both \(x\) and \(y\), is a circle. By plotting these figures on the same graph, you can observe intersection points where their paths cross.
These graphical intersections directly correspond to solutions that satisfy both equations in the system. Therefore, understanding graphical solutions can provide insight into the nature and number of solutions, whether they are real or complex numbers, and how they relate to one another on the coordinate plane.

Factoring Equations for Solutions

Factoring is a key technique used to solve quadratic equations. It involves breaking down a more complex expression into simpler factors that, when multiplied together, give the original equation. This technique can simplify finding solutions, especially when dealing with polynomials.
In the case of the given exercise, after substituting the second equation into the first, you arrive at the expression \(y^4 - 3y^2 = 0\). This can be factored into \(y^2(y^2 - 3) = 0\).
The zeros of this factored equation give possible solutions for \(y\):

• \(y^2 = 0\) which means \(y = 0\).
• \(y^2 - 3 = 0\) which simplifies to \(y = \pm \sqrt{3}\).

Having identified these solutions for \(y\), we then find corresponding \(x\) values using the original substitution. Factoring thus transforms what could be a complex problem into straightforward steps, making it easier to identify the points of intersection for the system of equations.