Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Appendix A - Equations and Inequalities - Page 420: 48

Answer

{$\frac{-1\pm2\sqrt 5}{4}$}

Work Step by Step

$(4x+1)^{2}=20$ Using generalized square root property, $4x+1=\pm\sqrt (20)$ $4x+1=\pm\sqrt (4\times5)$ $4x+1=\pm\sqrt 4\sqrt 5$ $4x+1=\pm2\sqrt 5$ Subtracting 1 from both sides, $4x+1-1=-1\pm2\sqrt 5$ $4x=-1\pm2\sqrt 5$ Dividing both sides by 4, $\frac{4x}{4}=\frac{-1\pm2\sqrt 5}{4}$ $x=\frac{-1\pm2\sqrt 5}{4}$ Check: $(4(\frac{-1+2\sqrt 5}{4})+1)^{2}=20$ $(-1+2\sqrt 5+1)^{2}=20$ $(2\sqrt 5)^{2}=20$ $4(\sqrt 5)^{2}=20$ $4(5)=20$ $20=20$ $(4(\frac{-1-2\sqrt 5}{4})+1)^{2}=20$ $(-1-2\sqrt 5+1)^{2}=20$ $(-2\sqrt 5)^{2}=20$ $4(\sqrt 5)^{2}=20$ $4(5)=20$ $20=20$ Therefore, the solution set is {$\frac{-1\pm2\sqrt 5}{4}$}
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