Chapter 10, Problem 93RE

To determine

To find: The intersection points of the graph

Answer to Problem 93RE

The intersection points of the system of equations are $(2,-2)$ and $\left(-\frac{1}{2},\frac{7}{4}\right)$

Explanation of Solution

Given:

  $\begin{array}{l}{x}^2+y=2\\ x^2-3x-y=0\end{array}$

  

Calculation:

The given function is

  $x^2+y=2$ Equation $1$

  $x^2-3x-y=0$ Equation $2$

Solve the Equation $1$ for $y$

  $x^2+y=2$

  $y=2-x^2$

Substitute $y=2-x^2$ in $x^2-3x-y=0$ and solve for $x$

  $x^2-3x-(2-x^2)=0$ Substitute $y=2-x^2$

  $x^2-3x-2+x^2=0$

  $2x^2-3x-2=0$ Combined like terms

  $2x^2-4x+x-2=0$ Factorize

  $\begin{array}{l}2x(x-2)+1(x-2)=0\\ (x-2)(2x+1)=0\end{array}$

  $x=2$ Or $2x=-1$ apply zero product property

  $x=2$ Or $x=-\frac{1}{2}$

Substitute the values of $x$ in the equation $y=2-x^2$

For $x=2$

  $\begin{array}{l}$ y=2-{(2)}^2$=2-4\\ =-2\end{array}$

For $x=-\frac{1}{2}$

  $\begin{array}{l}y=2-{\left(-\frac{1}{2}\right)}^2\\ =2-\frac{1}{4}\\ =\frac{8-1}{4}\\ =\frac{7}{4}\end{array}$

Thus the two solutions are $(2,-2)$ and $\left(-\frac{1}{2},\frac{7}{4}\right)$

Conclusion:

Hence, the intersection points of the system of equations are $(2,-2)$ and $\left(-\frac{1}{2},\frac{7}{4}\right)$