Chapter 10.8, Problem 38E

To determine

The given system of equation has to be solved for its solutions by using the graphical method:

Answer to Problem 38E

The set of equations $\{\begin{array}{l}{x}^2-y^2=3\\ y=x^2-2x-8\end{array}$ has four solutions by using the graphical method:

  $\begin{array}{l}\left(-1.88,-0.72\right)\\ \left(4.65,4.31\right)\\ \left(3.45,-2.99\right)\\ \left(-2.22,1.4\right)\end{array}$

  $\{\begin{array}{l}{x}^2-y^2=3\\ y=x^2-2x-8\end{array}$

Explanation of Solution

Given:

  $\begin{array}{l}{x}^2-y^2=3\\ y=x^2-2x-8\end{array}$

Concept Used:

When the quadratic equation’s graph crosses the $x$ -axis at two points then it is said to be the equation system has two solutions.

When the quadratic equation’s graph touches the $x$ -axis at a single point then it is said to be the equation system has one solution.

There is no real solution for equation system if the graph does not intersect with the $x$ -axis.

The slope intercept form is given as:

  $y=mx+b$

Where,

Slope: $m$

  $y-$ intercept: $b$

Calculations:

The given eqns. are:

  $x^2-y^2=3$........(1)

  $y=x^2-2x-8$.........(2)

Need to write the given equations in the form of the slope-interceot form:

From eq. (1):

  $\begin{array}{l}{x}^2-y^2=3\\ y^2=x^2-3\end{array}$

  $y=\sqrt{x^2-3}$.......(3)

Eq. (2) and (3) graphs are drawn as below:

  

From the graph:

The solution set of the given equation is:

  $\begin{array}{l}\left(-1.88,-0.72\right)\\ \left(4.65,4.31\right)\\ \left(3.45,-2.99\right)\\ \left(-2.22,1.4\right)\end{array}$

Conclusion:

Hence, the given set of equations has four solutions which are well labelled in the above graph.