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

1 Answer
Nghi N.
Sep 16, 2016

#2 +- sqrt39/3#

Explanation:

Factor the trinomial:
#(x^2 - 3x - 4) = (x + 1)(x - 4)#
After simplifying both sides by (x + 1), the equation becomes;
#1/(x - 4) = 3x#
3x(x - 4) - 1 = 0
#3x^2 - 12x - 1 = 0#
#D = d^2 = b^2 - 4ac = 144 + 12 = 156# --> #d = +-2sqrt39#
There are 2 real roots:
#x = -b/(2a) +- d/(2a) = 12/6 +- (2sqrt39)/6 = 2 +- sqrt39/3#

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