Chapter 8 - Section 8.6 - Systems of Nonlinear Equations - 8.6 Assess Your Understanding - Page 616: 73

$b=\pm2$ and $a=\pm2$

the product of two numbers, $ab=4$ and the sum of their squares, $a^2+b^2=8,$ Thus, $a=\frac{4}{b},$ $(\frac{4}{b})^2+b^2=8,$ $\frac{16}{b^2}+b^2=8,$ $16+b^4=8b^2,$ $b^4-8b^2+16=0,$ let $b^2=k$ $k^2-8k+16=0,$ $k^2-4k-4k+16=0,$ $k(k-4)-4(k-4)=0,$ $(k-4)(k-4)=0,$ Thus, $(b^2-4)(b^2-4)=0,$ $(b-2)(b+2)(b-2)(b+2)=0,$ $b=\pm2$ and $a=\pm2$