How do you solve for x in the simplest radical form for #(x^2+2x+4)/x =(2x)/1#?

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Answer 1 · 2015-04-14T10:32:49

If #a/b=c/d->a*d=b*c# this is called cross-product.

Applying this, we have:

#(x^2+2x+4)*1=x*2x->#

#x^2+2x+4=2x^2->#

Subtract #2x# on both sides and multiply by #-1#

#x^2-2x-4=0#

There are several ways to solve this. I will rewrite for the square:

#(x^2-2x+1)-5=0->#
#(x-1)^2-5=0->#
#(x-1)^2=5->#
#x-1=+sqrt5or-sqrt5->#

Solution:

#x=1+sqrt5orx=1-sqrt5#