Chapter 7.1, Problem 49E

To determine

Solve the system graphically or algebraically. Explain your choice of method.

  $\{\begin{array}{c}x-2y=4\\ x^2-y=0\end{array}$

Answer to Problem 49E

The solution for the equations given is $x=4-\frac{15\pm \sqrt{31}i}{4},y=\frac{-15\pm \sqrt{31}i}{8}$

Explanation of Solution

Given:

The given equations are $\{\begin{array}{c}x-2y=4\\ x^2-y=0\end{array}$

Concept Used:

The given equations are in two variables x and y. These equations can be solved using graphical method and algebraic method. This equations are simple so they can be solved algebraically.

By substituting one equation in the other we can solve the equations and find x and y values.

Calculation:

Let us first substitute x from $x-2y=4$ as $x=4+2y$ in $x^2-y=0$

Now the equation changes to ${\left(4+2y\right)}^2-y=0$ after substituting x. By solving this we get y value.

  $\begin{array}{l}{\left(4+2y\right)}^2-y=0\\ \left(16+16y+4y^2\right)-y=0\\ 4y^2+15y+16=0\\ y=\frac{-15\pm \sqrt{31}i}{8}\end{array}$

We get it as an imaginary root and by substituting it we get an imaginary x value as

  $\begin{array}{l}x=4+2y\\ x=4+2\left(\frac{-15\pm \sqrt{31}i}{8}\right)\\ x=4-\frac{15\pm \sqrt{31}i}{4}\end{array}$

Conclusion:

The solution for the equations given is $x=4-\frac{15\pm \sqrt{31}i}{4},y=\frac{-15\pm \sqrt{31}i}{8}$