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Q. 17

Expert-verifiedFound in: Page 772

Book edition
6th

Author(s)
Sullivan

Pages
1200 pages

ISBN
9780321795465

In Problems 5–24, graph each equation of the system. Then solve the system to find the points of intersection.

${x}^{2}+{y}^{2}=4;\phantom{\rule{0ex}{0ex}}{y}^{2}-x=4$

The graph of the system of equations ${x}^{2}+{y}^{2}=4;{y}^{2}-x=4$ is:

The system of non-linear equation:

${x}^{2}+{y}^{2}=4;\phantom{\rule{0ex}{0ex}}{y}^{2}-x=4$

To graph the equation and to find the point of intersection.

Graph the equations in the same plane.

Substitute the second equation in the first equation,

${({y}^{2}-4)}^{2}+{y}^{2}=4;\phantom{\rule{0ex}{0ex}}{y}^{4}-8{y}^{2}+16+{y}^{2}=4\phantom{\rule{0ex}{0ex}}{y}^{4}-7{y}^{2}+12=0$

${y}^{4}-7{y}^{2}+12=0\phantom{\rule{0ex}{0ex}}({y}^{2}-4)({y}^{2}-3)=0\phantom{\rule{0ex}{0ex}}{y}^{2}=3,4\phantom{\rule{0ex}{0ex}}y=\pm \sqrt{3},\pm 2$

When $y=\pm \sqrt{3}$,

${x}^{2}+{(\pm \sqrt{3})}^{2}=4\phantom{\rule{0ex}{0ex}}{x}^{2}+3=4\phantom{\rule{0ex}{0ex}}{x}^{2}=1\phantom{\rule{0ex}{0ex}}x=\pm 1$

When $y=\pm 2$,

${x}^{2}+{(\pm 2)}^{2}=4\phantom{\rule{0ex}{0ex}}{x}^{2}+4=4\phantom{\rule{0ex}{0ex}}{x}^{2}=0\phantom{\rule{0ex}{0ex}}x=0$

The points of intersection are $(-1,-\sqrt{3})(-1,\sqrt{3}),(0,-2),(0,2)$

The graph of the system of non-linear equations with the point of intersection is:

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