Chapter 10, Problem 58SGA

To determine

To find: Each equation $$

  $x^2-4x-4y=4,{\left(x-2\right)}^2+4y=0$ algebraically and check the solutions by graphically.

Answer to Problem 58SGA

Therefore, the four solutions are $\left(-1.2,3.2\right),\left(1.2,-1.2\right),\left(-3.2,1.2\right),\left(3.2,-1.2\right)$ same as the solutions obtained from the graph.

Explanation of Solution

Given information:

The equation is,

  $\begin{array}{c}{x}^2+y^2=12\\ xy=-4\end{array}$

Consider the system of equations,

  $\begin{array}{c}{x}^2+y^2=\mathrm{12.....}(1)\\ xy=-\mathrm{4.....}(2)\end{array}$

Solve equation (2) for $x$ ,

  $\begin{array}{l}xy=-4\\ x=\frac{-4}{y}\mathrm{.....}(3)\end{array}$

Put the value of $x$ from equation (3) to equation (1),

  $\begin{array}{c}{x}^2+y^2=12\\ y^2+{\left(\frac{-4}{y}\right)}^2=12\\ y^4+16=12y^2\mathrm{......}(4)\end{array}$

Let,

  $y^2=t$

Substitute the value in equation (4).

  $\begin{array}{c}{y}^4+16=12y^2\\ t^2+16=12t\end{array}$

To make Zero on the right hand side add $-12t$ on both sides,

  $t^2-12t+16=0$ .

Use quadratic formula to find the value of $t$ .

  $\begin{array}{c}t=\frac{12\pm \sqrt{{\left(-12\right)}^2-4\cdot 1\cdot 16}}{2}\\ =\frac{12\pm \sqrt{80}}{2}\end{array}$

It gives two values of $t$ such as,

  $t=10.47,t=1.53$ .

Now replace $tby{y}^2$ .

  $\begin{array}{c}{y}^2=10.67\\ y=\pm 3.24\end{array}$

And,

  $\begin{array}{c}{y}^2=1.53\\ y=\pm 1.24\end{array}$

Substitute the value of $y$ in $\frac{y}{x}$ .

For $y=-3.24$ ,

  $\begin{array}{c}x=\frac{-4}{y}\\ =\frac{-4}{-3.24}\\ =1.23\end{array}$

For $x=3.24$ ,

  $\begin{array}{c}x=\frac{-4}{y}\\ =\frac{-4}{3.24}\\ =-1.23\end{array}$

For $x=1.24$ ,

  $\begin{array}{c}x=\frac{-4}{y}\\ =\frac{-4}{1.24}\\ =-3.23\end{array}$

For $x=-1.24$ ,

  $\begin{array}{c}x=\frac{-4}{y}\\ =\frac{-4}{-1.24}\\ =3.23\end{array}$

Therefore, the solutions are $\left(-1.2,3.2\right),\left(1.2,-1.2\right),\left(-3.2,1.2\right),\left(3.2,-1.2\right)$ .

The graph of the first equation is circle,

  

The second equation is hyperbola,

  

Therefore, the four solutions are $\left(-1.2,3.2\right),\left(1.2,-1.2\right),\left(-3.2,1.2\right),\left(3.2,-1.2\right)$ same as the solutions obtained from the graph.