Chapter 9, Problem 16CR

To determine

To find the number of solutions for the system $x^2-4y^2=4$ and $x+y^2=1$ .

Answer to Problem 16CR

The answer is $c.$

  $2$ .

Explanation of Solution

Given:

Equations:

  $x^2-4y^2=4$

  $x+y^2=1$

Calculation:

Given Equations:

  $x^2-4y^2=4$

  $x+y^2=1$

Solving equation, $x+y^2=1$ for value of $x$ , we have:

  $x=1-y^2$

Putting $x=1-y^2$ in equation $x^2-4y^2=4$ , we have:

  $\begin{array}{l}{\left(1-y^2\right)}^2-4y^2=4\\ 1+y^4-2y^2-4y^2-4=0\\ y^4-6y^2-3=0\\ y=-\sqrt{\left(3-2\sqrt{3}\right)},\sqrt{\left(3-2\sqrt{3}\right)}\end{array}$

Thus, when $y=-\sqrt{\left(3-2\sqrt{3}\right)},\sqrt{\left(3-2\sqrt{3}\right)}$

  $x=2\left(\sqrt{3}-1\right),2\left(\sqrt{3}-1\right)$

Thus, there are two solutions to the given system.

Conclusion:

The answer is $c.$ $2$ .