Problem 60
Question
Find any points of intersection of the graphs of the equations algebraically and then verify using a graphing utility. $$\begin{aligned} &-4 x^{2}+9 y^{2}-36 y=0\\\ &x^{2}+9 y-27=0 \end{aligned}$$
Step-by-Step Solution
Verified Answer
The solutions are \( ( x, y) = (\pm 3\sqrt{2},1), (\pm 3\sqrt{6},2)\)
1Step 1: Rewriting for clarity
Let's start by rewriting the equations. The given equation has two equations which can be rewritten as \(4x^2 + 36y - 9y^2 = 0\) and \(x^2 + 9y -27 = 0\). The goal is to find \(x\) and \(y\) such that they satisfy both equations.
2Step 2: Equate for algebraic solution
Equating the two equations together gives \(4x^2 + 36y - 9y^2 = x^2 + 9y -27\). Solving for \(x\) and \(y\), we move all terms to one side to obtain \(3x^2 - 9y^2 + 63 = 0\).
3Step 3: Solve for x and y
By factoring, we find that there are two possible solutions for \(x\), namely \(x = 3 \sqrt{3y^2 - 7}\) or \(x = - 3 \sqrt{3y^2 - 7}\). Substituting \(y\) from second equation into the solutions, we get four sets of solutions.
Key Concepts
System of Quadratic EquationsFactoring Quadratic EquationsGraphing Utility Verification
System of Quadratic Equations
Often in algebra, we come across a system of quadratic equations, which comprises two or more equations involving the same variables. To solve a system, we look for the values of the variables that satisfy all equations simultaneously.
To handle this algebraically, one common method is substitution, where we solve one equation for one variable and substitute the result into the other equation. Another method is elimination, where we add or subtract equations to eliminate one variable, making it easier to solve for the remaining one. The intersection points of the graphs represent the solutions to the system, and each one corresponds to an instance where both equations are true.
In the case of our exercise, the equations given are quadratic, meaning they each contain terms up to the second degree. The intersection points of their respective graphs in the coordinate plane will give us the solutions. When dealing with quadratics, it's not uncommon to have multiple intersection points due to the curved nature of their graphs.
To handle this algebraically, one common method is substitution, where we solve one equation for one variable and substitute the result into the other equation. Another method is elimination, where we add or subtract equations to eliminate one variable, making it easier to solve for the remaining one. The intersection points of the graphs represent the solutions to the system, and each one corresponds to an instance where both equations are true.
In the case of our exercise, the equations given are quadratic, meaning they each contain terms up to the second degree. The intersection points of their respective graphs in the coordinate plane will give us the solutions. When dealing with quadratics, it's not uncommon to have multiple intersection points due to the curved nature of their graphs.
Factoring Quadratic Equations
Factoring is a powerful method used to solve quadratic equations, which are polynomials of the second degree typically written in the form of \( ax^2 + bx + c = 0 \). The factoring process involves breaking down the quadratic equation into a product of simpler expressions that are easier to solve.
There are several methods for factoring quadratics, including using the quadratic formula, completing the square, or looking for two binomials that multiply to give the original quadratic. When the quadratic is set to zero, factoring enables us to apply the zero-product property which states if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \).
In our exercise, factoring the equation \( 3x^2 - 9y^2 + 63 = 0 \) could reveal potential solutions for \( x \) and \( y \). This is a critical step as it simplifies complex equations, making it possible to solve for the variables involved.
There are several methods for factoring quadratics, including using the quadratic formula, completing the square, or looking for two binomials that multiply to give the original quadratic. When the quadratic is set to zero, factoring enables us to apply the zero-product property which states if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \).
In our exercise, factoring the equation \( 3x^2 - 9y^2 + 63 = 0 \) could reveal potential solutions for \( x \) and \( y \). This is a critical step as it simplifies complex equations, making it possible to solve for the variables involved.
Graphing Utility Verification
Verification with a graphing utility is the process of confirming the solutions we've found algebraically. After solving equations on paper, we can use a graphing utility, such as a software program or a graphing calculator, to graph the functions and visually check if the algebraic solutions correspond to the intersection points on the graphs.
These tools are incredibly helpful for double-checking work and also for providing a visual representation that often makes understanding the nature of the solutions easier. For example, if we were to enter the equations from our exercise into a graphing utility, we would expect to see the graphs intersecting at the points that correspond to our algebraically derived solutions.
Moreover, a graphing utility can also reveal additional insights, such as the number of intersection points or whether the graphs are symmetrical, which might not be immediately apparent from the algebraic method alone. Therefore, combining algebraic techniques with graphical verification provides a robust approach to solving mathematical problems.
These tools are incredibly helpful for double-checking work and also for providing a visual representation that often makes understanding the nature of the solutions easier. For example, if we were to enter the equations from our exercise into a graphing utility, we would expect to see the graphs intersecting at the points that correspond to our algebraically derived solutions.
Moreover, a graphing utility can also reveal additional insights, such as the number of intersection points or whether the graphs are symmetrical, which might not be immediately apparent from the algebraic method alone. Therefore, combining algebraic techniques with graphical verification provides a robust approach to solving mathematical problems.
Other exercises in this chapter
Problem 60
Use the results of Exercises \(49-52\) to find a set of parametric equations to represent the graph of the line or conic. Hyperbola: vertices: (±2,0)\(;\) foci:
View solution Problem 60
Find the distance between the point and the line. $$\begin{array}{cc}\text{Point} && \text{Line} \\ (2,1) && y=x+2\end{array}$$
View solution Problem 60
Find the equation of the tangent line to the parabola at the given point. $$y=-2 x^{2},(2,-8)$$
View solution Problem 60
Classify the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola. $$4 x^{2}+25 y^{2}+16 x+250 y+541=0$$
View solution