Chapter 9.8, Problem 18WE

To determine

To calculate: To find the real solutions

Answer to Problem 18WE

Solutions are $\left(\frac{3}{\sqrt{2}},\sqrt{2}\right),\left(-\frac{3}{\sqrt{2}},-\sqrt{2}\right)$

Explanation of Solution

Given information: System of equations $2x^2-y^2=7;xy=3$

Calculation:

Given the system of equations as follows:

  $\begin{array}{l}2x^2-y^2=7----(1)\\ xy=3-------(2)\end{array}$

From equation (2)

  $\begin{array}{l}xy=3\\ y=\frac{3}{x}\end{array}$

Substituting the value in equation (1)

  $\begin{array}{l}2x^2-{\left(\frac{3}{x}\right)}^2=7\\ 2x^2-\frac{9}{x^2}=7\\ 2x^4-9=7x^2\\ 2x^4-7x^2-9=0\\ 2x^4-9x^2+2x^2-9=0\\ x^2\left(2x^2-9\right)+1\left(2x^2-9\right)=0\\ \left(2x^2-9\right)\left(x^2+1\right)=0\\ x^2=\frac{9}{2},-1\\ x=\pm \frac{3}{\sqrt{2}}\end{array}$

When $x=\frac{3}{\sqrt{2}}$

  $y=\frac{3}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}=\sqrt{2}$

When $x=-\frac{3}{\sqrt{2}}$

  $y=\frac{3}{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}=-\sqrt{2}$

Conclusion:

Hence, solutions are $\left(\frac{3}{\sqrt{2}},\sqrt{2}\right),\left(-\frac{3}{\sqrt{2}},-\sqrt{2}\right)$