## Trigonometry (11th Edition) Clone

{$\frac{-1\pm2\sqrt 5}{4}$}
$(4x+1)^{2}=20$ 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 {$\frac{-1\pm2\sqrt 5}{4}$}