Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 9 Quadratic Relations and Conic Sections - Chapter Review - Page 672: 35



Work Step by Step

$y^2=4x\\0.5y^2=2x$ Thus plugging this into the second equation we get: $0.5y^2-5y+8=0\\y^2-10y+16=0\\(y-2)(y-8)=0$ Thus $y=2$ or $y=8$. $x=0.25y^2$. Thus if $y=2$, then $x=1$, and if $y=8$, then $x=16$. Thus the solutions are: $(1,2),(16,8)$
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