Chapter 8.2, Problem 63E

Solution

To solve: the system of equations both algebraically and graphically.

The value of intersection points are

  $\begin{array}{l}\left(y=\pm \sqrt{\frac{\sqrt{161}+47}{32}},-\frac{1+\sqrt{161}}{16}\right)\\ \left(\pm \sqrt{\frac{47-\sqrt{161}}{32}},\frac{-1-\sqrt{161}}{16}\right)\end{array}$ .

Given information: The given system of an equations is:

  $\begin{array}{c}{x}^2+4y^2=4\\ y=2x^2-3\end{array}$

Calculation:

The value of $x^2$

can be calculated from an equation (2):

  $\begin{array}{c}{x}^2+4y^2=4\mathrm{........}(1)\\ y=2x^2-3\\ y+3=2x^2\\ \frac{y+3}{2}=x^2\mathrm{........}(2)\end{array}$

Substitute the value of equation (2) in equation (1):

  $\begin{array}{c}\frac{y+3}{2}+4y^2=4\\ \frac{y+3+8y^2}{2}=4\\ y+3+8y^2=8\\ y+3+8y^2-8=0\\ 8y^2+y-5=0\end{array}$

Using the Quadratic formulae:

  $\begin{array}{l}y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ y=\frac{-1\pm \sqrt{1-4\times (-5)\times 8}}{2\times 8}\\ y=\frac{-1\pm \sqrt{1+160}}{16}\\ y=\frac{-1\pm \sqrt{161}}{16}\\ y=\frac{-1+\sqrt{161}}{16},y=\frac{-1-\sqrt{161}}{16}\end{array}$

If the value of $y=\frac{-1+\sqrt{161}}{16}$ then the value of,

  $\begin{array}{l}{x}^2=\frac{\frac{-1+\sqrt{161}}{16}+3}{2}\\ x^2=\frac{-1+\sqrt{161}+48}{32}\\ x^2=\frac{-1+\sqrt{161}+48}{32}\\ x^2=\frac{\sqrt{161}+47}{32}\\ x=\pm \sqrt{\frac{\sqrt{161}+47}{32}}\end{array}$

If the value of $y=\frac{-1-\sqrt{161}}{16}$ , then the value of

  $\begin{array}{l}{x}^2=\sqrt{\frac{\frac{-1-\sqrt{161}}{16}+3}{2}}\\ x^2=\frac{-1-\sqrt{161}+48}{32}\\ x^2=\frac{47-\sqrt{161}}{32}\\ x^2=\frac{-\sqrt{161}+47}{32}\\ x=\pm \sqrt{\frac{47-\sqrt{161}}{32}}\end{array}$

Thus, the intersection points are

  $\left(\pm \sqrt{\frac{\sqrt{161+47}}{32}},\frac{-1+\sqrt{161}}{16}\right)\left(\pm \sqrt{\frac{47-\sqrt{161}}{32}},\frac{-1-\sqrt{161}}{16}\right)$

Which are equivalent to $\left(\pm 1.3666,0.731\right)$ and $\left(\pm 1.035,-0.856\right)$ .

Here, the graphical representation is as follow:

  

Graphical Interpretation:

In the graph the red curve represents $x^2+4y^2=4$ and blue curve represents $y=2x^2-3$ .And from the graph it is observed that the curves are intersecting at 4 points, $\left(\pm 1.3666,0.731\right)$ and $\left(\pm 1.035,-0.856\right)$ .