How do you solve the quadratic #4-1/x=3/x^2# using any method?

1 Answer
Dec 27, 2016

Get rid of the denominators by multiplying the whole equation by #color(blue)(x^2)#

#4 color(blue)( xx x^2) - 1/x color(blue)(xx x^2) = 3/x^2 color(blue)(xx x^2)#

#4x^2 -x = 3" "larr# make equal to 0

#4x^2 -x-3 =0#

Find factors of 4 and 3 which subtract to give 1.

#(4x+3)(x-1) =0" "larr# either factor could be =0

If #4x+3 = 0" "rarr 4x=-3" "rarr x = -3/4#

If#x-1=0" " x = 1#