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Chapter 8: Problem 9

Solve the given system of nonlinear equations. Use a graph to help you avoid any potential extraneous solutions. $$ \left\\{\begin{aligned} x+2 y^{2} &=2 \\ x^{2}+4 y^{2} &=4 \end{aligned}\right. $$

Short Answer

Expert verified
The solutions are \((0, 1)\) and \((0, -1)\).

Step by step solution

01

Graph the equations

First, rewrite each equation in a form that can be graphed easily.\- For the first equation, resolve for \(x\): \(x = 2 - 2y^2\).\- For the second equation, rearrange it to match a standard form: \(x^2 = 4 - 4y^2\).\- Identify these as a parabola opening left/right and an ellipse centered at the origin, respectively, and then graph them.
02

Determine the intersections graphically

Using the graph, locate the points where the parabola and the ellipse intersect. These points are solutions to the system of equations, and finding them graphically helps avoid extraneous solutions that might occur with purely algebraic manipulation.
03

Solve the system algebraically

To confirm the graphical intersections, begin solving algebraically. Substitute the expression for \(x\) from the first equation into the second equation:\[ (2 - 2y^2)^2 + 4y^2 = 4 \].
04

Simplify and solve the resulting equation

Expand \((2 - 2y^2)^2\) to get: \(4 - 8y^2 + 4y^4\).Combine terms to form a polynomial: \[ 4y^4 - 12y^2 + 4 = 0 \].Factor this as: \( (2y^2 - 2)(2y^2 - 2) = 0 \), leading to \(y^2 = 1\). So, \(y = \pm 1\).
05

Find corresponding x-values

For each \(y\) value, substitute back into the first equation to find \(x\):\- For \(y = 1\), \(x = 2 - 2(1^2) = 0\).\- For \(y = -1\), \(x = 2 - 2(-1)^2 = 0\).
06

Compile the solutions

The solutions from the steps above are the pairs: \((0, 1)\) and \((0, -1)\). These points satisfy both original equations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Graphical Intersection
Graphical intersection is a helpful method for solving systems of equations, especially nonlinear ones. When dealing with these systems, you often get curves that aren't simply straight lines, like parabolas and ellipses. By graphing, you can visually pinpoint where these curves cross. These crossing points, or intersections, reveal the solutions to the system.

Using graphs,
  • You can see all potential solutions at a glance.
  • Avoid the risk of extraneous solutions that sometimes appear when solving only with algebra.
  • Confirm algebraic results with a visual check.
In our problem, the graphs of a parabola and an ellipse were used to find intersection points, which represented the solutions: (0, 1) and (0, -1). By plotting these curves and observing their intersections, it provided a guide for further algebraic steps.
Parabola
A parabola is a symmetric curve that can open upwards, downwards, or sideways. It is commonly represented by a quadratic equation such as the rearranged version of the first equation: \[ x = 2 - 2y^2 \] This equation describes a parabola

  • Opening to the left.
  • Centered around the line \(y\)-axis.
A parabola's distinctive u-shape depends on the orientation of the quadratic term. In this exercise, the parabola's equation was manipulated to assist in visual understanding of the graphical intersection. When graphed alongside the ellipse, their intersection provided crucial solutions.
Ellipse
Ellipses are elongated circles represented by quadratic equations where the x and y terms have scaled values. The equation from our problem: \[ x^2 = 4 - 4y^2 \] This is an equation for an ellipse centered at the origin,
  • Extended equally horizontally and vertically when rearranged.
  • Containing symmetric axes, simplifying the graphing process.
Ellipses differ from circles mainly by having two axes of symmetry, which makes them crucial in understanding nonlinear intersections. By evaluating where this ellipse intersected with the parabola, we found our systems' solutions.
Algebraic Manipulation
Algebraic manipulation involves transforming equations to simplify or solve them further. This can include rearranging, substituting, or factoring equations to find solutions.

In this problem, algebraic manipulation helped confirm which intersections were legitimate solutions. By solving algebraically, questions about accuracy and extraneous solutions were addressed.
  • Substitution: Express \(x\) from the first equation and replace it in the second.
  • Expansion: Convert expressions like \((2 - 2y^2)^2\) into standard form through expansion.
  • Factoring: Simplify to \((2y^2 - 2)^2 = 0\) to find \(y\) solutions.
Completing these steps ensured clear, reliable answers that matched the graphical intersections previously identified.

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Most popular questions from this chapter

Solve the given system of nonlinear equations. Sketch the graph of both equations on the same set of axes to verify the solution set. $$ \left\\{\begin{aligned} x^{2}+y^{2} &=16 \\ \frac{1}{9} y^{2}-\frac{1}{16} x^{2} &=1 \end{aligned}\right. $$

Consider the following scenario. In the small village of Pedimaxus in the country of Sasquatchia, all 150 residents get one of the two local newspapers. Market research has shown that in any given week, \(90 \%\) of those who subscribe to the Pedimaxus Tribune want to keep getting it, but \(10 \%\) want to switch to the Sasquatchia Picayune. Of those who receive the Picayune, \(80 \%\) want to continue with it and \(20 \%\) want switch to the Tribune. We can express this situation using matrices. Specifically, let \(X\) be the 'state matrix' given by $$ X=\left[\begin{array}{l} T \\ P \end{array}\right] $$ where \(T\) is the number of people who get the Tribune and \(P\) is the number of people who get the Picayune in a given week. Let \(Q\) be the 'transition matrix' given by $$ Q=\left[\begin{array}{ll} 0.90 & 0.20 \\ 0.10 & 0.80 \end{array}\right] $$ such that \(Q X\) will be the state matrix for the next week. Show that \(S Y=X_{s}\) for any matrix \(Y\) of the form $$ Y=\left[\begin{array}{r} y \\ 150-y \end{array}\right] $$ This means that no matter how the distribution starts in Pedimaxus, if \(Q\) is applied often enough, we always end up with 100 people getting the Tribune and 50 people getting the Picayune.

As we stated at the beginning of this section, the technique of resolving a rational function into partial fractions is a skill needed for Calculus. However, we hope to have shown you that it is worth doing if, for no other reason, it reinforces a hefty amount of algebra. One of the common algebraic errors the authors find students make is something along the lines of $$ \frac{8}{x^{2}-9} \neq \frac{8}{x^{2}}-\frac{8}{9} $$ Think about why if the above were true, this section would have no need to exist.

Compute the determinant of the given matrix. (Some of these matrices appeared in Exercises \(1-8\) in Section 8.4.) \(H=\left[\begin{array}{rrrr}1 & 0 & -3 & 0 \\ 2 & -2 & 8 & 7 \\ -5 & 0 & 16 & 0 \\ 1 & 0 & 4 & 1\end{array}\right]\)

Compute the determinant of the given matrix. (Some of these matrices appeared in Exercises \(1-8\) in Section 8.4.) \(V=\left[\begin{array}{rrr}i & j & k \\ -1 & 0 & 5 \\ 9 & -4 & -2\end{array}\right]\)
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