I tried to solve the following equation but I got -6 under the radical sign... Here´s the equation I tried to solve: (y-4)^2 = 64 Any help perhaps?
Last edited by Bluerabbits; Apr 19th 2011 at 10:08 AM.
Follow Math Help Forum on Facebook and Google+
Originally Posted by Bluerabbits I tried to solve the following equation but I got -6 under the radical sign... Here´s the equation I tried to solve: (y-4)^2 = 64 Any help perhaps? Take the square root of both sides: y - 4 = +/- sqrt(64) y - 4 = +/-8 Voila! No more radicals! -Dan
Or... $\displaystyle y^2$ - 8y + 16 = 64 $\displaystyle \Rightarrow$ $\displaystyle y^2$ - 8y - 48 = 0 $\displaystyle \Delta$=256, $\displaystyle y_1$ and $\displaystyle y_2$ = ...
Originally Posted by topsquark Take the square root of both sides: y - 4 = +/- sqrt(64) y - 4 = +/-8 Voila! No more radicals! -Dan Thank you! answers: y1 = 12 y2 = -4 - Anika
View Tag Cloud