How do you solve #\frac { 1} { x } = \frac { x + 1} { 6}#?

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Answer 1 · 2016-11-06T12:01:36

#x=2,# or #x=-3#

#1/x=(x+1)/6#; #6=x^2+x#

Thus #x^2+x-6=0#

#x^2+x=6#

#x^2+x+1/4=6+1/4=25/4#

#(x+1/2)^2=25/4#

#x+1/2=+-sqrt(25/4)#

#x=+-5/2-1/2#; i.e. #x=2,# or #x=-3#

Here, I completed a square, which I am sure that you covered in your course. Look at the appropriate section in your text for more explanation.