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

1 Answer
Apr 14, 2015

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#