Chapter 5.4, Problem 51E

To determine

To solve: The given non-linear system. Also, justify the answer with a graph.

Answer to Problem 51E

The solution is: $\text{(0,}-\text{1)}$ .

Explanation of Solution

Given information:

The following equations:

  $\begin{array}{l}{x}^2+y^2=1\\ y=\frac{1}{2}{x}^2-1\end{array}$

Formula used:

The following expansion formula:

  ${(a-b)}^2=a^2-2ab+b^2$

Calculation:

Substituting the value $y=\frac{1}{2}{x}^2-1$ in the equation $x^2+y^2=1$ ,

  $\begin{array}{l}{x}^2+y^2=1[\text{Given]}\\ x^2+{\left(\frac{1}{2}{x}^2-1\right)}^2=1\left[\text{Substitute }\left(\frac{1}{2}{x}^2-1\right)\text{ for }y\right]\\ x^2+\frac{1}{4}{x}^4-x^2+1=1[\text{Expand]}\\ \frac{4x^2+x^4-4x^2+4}{4}=1[\text{Add fractions]}\\ 4x^2+x^4-4x^2+4=4[\text{Multiply both sides by 4]}\\ x^4+4=4[\text{Add like terms]}\\ x^4=\text{0 }[\text{Subtract 4 from both sides]}\\ x=0[\text{Raise each side to one-fourths]}\end{array}$

Substituting $x=0$ in the equation $y=\frac{1}{2}{x}^2-1$ ,

  $\begin{array}{l}y=\frac{1}{2}{x}^2-1[\text{Given]}\\ y=-\frac{3}{4}\cdot 0^2-1\text{[Substitute 0 for }x]\\ y=0-1[\text{Simplify]}\\ y=-1[\text{Subtract]}\end{array}$

Therefore, the solution obtained is, $\text{(0,}-1\text{)}$ .

It is verified on the graph as follows:

Graph:

Sketch the graph using graphing utility.

Step 1: Press WINDOW button to access the Window editor.

Step 2: Enter the equations $x^2+y^2=1$ and $y=\frac{1}{2}{x}^2-1$ which is required to graph.

Step 3: Press GRAPH button to graph the function.

The graph is obtained as:

  

Using the $\boxed{\text{INTERSECT}}$ feature on the graphing calculator obtain the points of intersection.

Interpretation :

The points of intersection obtained are the solutions of the equations. Since the solutions obtained by graph and by calculation are the same, the solutions are verified.