i swear i'll never understand problems with radicals >.<
anyway: $\displaystyle \sqrt{\sqrt{x+8}+x}=2$
show work if you can..someday i'll understand those things -.-..always confudling me..
i swear i'll never understand problems with radicals >.<
anyway: $\displaystyle \sqrt{\sqrt{x+8}+x}=2$
show work if you can..someday i'll understand those things -.-..always confudling me..
square both sides, you get
$\displaystyle \sqrt{x + 8} + x = 4$
$\displaystyle \Rightarrow \sqrt{x + 8} = 4 - x$
now square both sides again and continue
(and please, do not make the mistake of thinking $\displaystyle (4 - x)^2 = 4^2 - x^2$, it is wrong and will tick your professor off)
you're right so far
quadratic formula not needed. the quadratic was easily factorable$\displaystyle \frac{9\pm\sqrt{81-32}}{2}$ $\displaystyle \frac{9\pm7}{2}=4.5\pm3.5$
yes, x = 8 or x = 1 are the solutions to the quadratic. are both solutions valid for our original problem though?thus$\displaystyle x=1,8$?
also, that wasn't so hard, was it?