Chapter 9, Problem 13CT

To determine

To solve the system - $x^2-4y^2=4$ and $y^2=x+2$

Answer to Problem 13CT

The solutions are: $\left(6,2\sqrt{2}\right)$ , $\left(6,-2\sqrt{2}\right)$ and $\left(-2,0\right)$ .

Explanation of Solution

Given:

Equations: $x^2-4y^2=4$ and $y^2=x+2$ .

Calculation:

Given:

Equations: $x^2-4y^2=4$ and $y^2=x+2$ .

Putting the value of $y^2=x+2$ in equation $x^2-4y^2=4$ , we have:

  $\begin{array}{l}{x}^2-4y^2=4\\ x^2-4\left(x+2\right)=4\\ x^2-4x-8-4=0\\ x^2-4x-12=0\\ x^2-6x+2x-12=0\\ x\left(x-6\right)+2\left(x-6\right)=0\\ \left(x-6\right)\left(x+2\right)=0\end{array}$

Thus, $x=6,-2$

Putting the values of $x$ in $y^2=x+2$ , we have:

When $x=6$

  $\begin{array}{l}{y}^2=6+2\\ y^2=8\\ y=\pm 2\sqrt{2}\end{array}$

And when $x=-2$

  $\begin{array}{l}{y}^2=-2+2\\ y^2=0\\ y=0\end{array}$

Thus, the solutions are: $\left(6,2\sqrt{2}\right)$ , $\left(6,-2\sqrt{2}\right)$ and $\left(-2,0\right)$ .

Thus, the solutions are: $\left(6,2\sqrt{2}\right)$ , $\left(6,-2\sqrt{2}\right)$ and $\left(-2,0\right)$ .

Conclusion:

The solutions are: $\left(6,2\sqrt{2}\right)$ , $\left(6,-2\sqrt{2}\right)$ and $\left(-2,0\right)$ .