Q. 30 - Precalculus Enhanced with Graphing Utilities

Question

Graph each equation of the system. Then solve the system to find the points of intersection.

$\left\{\begin{array}{l}2x^2-xy+y^2=8\\ xy=4\end{array}\right.$

Step-by-Step Solution

Answer

The point of intersection is $(2,2),(-2,-2),(\sqrt{2},2\sqrt{2}),(-\sqrt{2},-2\sqrt{2})$.

Step 1. Given information

$\left\{\begin{array}{l}2x^2-xy+y^2=8\\ xy=4\end{array}\right.$

Step 2. Graphing each equation

$2x^2-xy+y^2=8$

$xy=4$

Step 3. Calculation of intersection point

$$\begin{gathered} Consider, \\ 2x^2-xy+y^2=8 .....\left(1\right) \\ xy=4 .....\left(2\right) \\ Now \\ y=\frac{4}{x} \\ substituting the value of y in eq.1 we get \\ 2x^2-x\left(\frac{4}{x}\right)+{\left(\frac{4}{x}\right)}^2=8 \\ 2x^2-4+\frac{16}{x^2}=8 \\ 2x^4+16=12x^2 \\ {x}^4-6x^2+8=0 \\ (x^2-4)(x^2-2)=0 \\ x=\pm 2,\pm \sqrt{2} \\ Now substituting value of x in eq. 2 we get \\ y=\pm \frac{4}{2} \\ =\pm 2 \\ y=\pm \frac{4}{\sqrt{2}} \\ =\pm 2\sqrt{2} \end{gathered}$$