SOLUTION: sqrt(x+4) = x^2-5 I squared both sides and eventually came out with x^4-10x^2-x+25. I got as far as x(x^3-10x-1) + 21 but I think I'm off base. Thank you so very much for all

Algebra ->  Square-cubic-other-roots -> SOLUTION: sqrt(x+4) = x^2-5 I squared both sides and eventually came out with x^4-10x^2-x+25. I got as far as x(x^3-10x-1) + 21 but I think I'm off base. Thank you so very much for all       Log On


   

Question 31950This question is from textbook College Algebra
: sqrt(x+4) = x^2-5 I squared both sides and eventually came out with x^4-10x^2-x+25. I got as far as x(x^3-10x-1) + 21 but I think I'm off base.
Thank you so very much for all your help in this madness! This question is from textbook College Algebra

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt(x+4)=x^2-5
Square both sides to get:
x+4=x^4-10x^2+25
x^4-10x^2-x+21=0
Graphing this with a TI-83
I get the following zeroes
x=2.7566944...
x=1.6214327...
x=-1.882801...
x=-2.495326...
Cheers,
Stan H.