Chapter 13 - Section 13.3 - Solving Nonlinear Systems of Equations - Exercise Set - Page 948: 32

${(4, -2), (4,2)}$

$y=\sqrt x$ $x^2+y^2=20$ $y=\sqrt x$ $y^2=(\sqrt x)^2$ $y^2=x$ $x^2+y^2=20$ $x^2+x=20$ $x^2+x−20=20−20$ $x^2+x−20=0$ $(x+5)(x−4)=0$ $x+5=0$ $x+5−5=0−5$ $x=−5$ $x−4=0$ $x−4+4=0+4$ $x=4$ $x=−5$ $y=\sqrt x$ $y=\sqrt {-5}$ Since we want the square root of a negative number, we find that x cannot equal −5. $x=4$ $y=\sqrt x$ $y=\sqrt 4$ $y=±2$