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
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-> 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
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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
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.
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