How do you solve #2/(x+1)=(3x)/4#?

1 Answer
Mar 4, 2018

#x=-1/2+sqrt(105)/5color(white)("xxx")"or"color(white)("xxx")x=-1/2-sqrt(105)/6#

Explanation:

Given
#color(white)("XXX")2/(x+1)=(3x)/4#

#rArr#
#color(white)("XXX")2xx4=(3x) xx (x+1)#

#color(white)("XXX")8=3x^2+3x#

#color(white)("XXX")3x^2+3x-8=0#

(applying the quadratic formula)
#color(white)("XXX")x=(-3+-sqrt(3^2-4 * 3 * (-8)))/(2 * 3)#

#color(white)("XXXx")=(-3+-sqrt(9+96))/6#

#color(white)("XXXx")=-1/2+-sqrt(105)/6#