How do you solve #5/(x^2+x-6) = 2 + (x-3)/(x-2)# and find any extraneous solutions?

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Answer 1 · 2017-07-11T17:23:26

#color(blue)(x=-1/3+sqrt(79)/3or x=-1/3-sqrt(79)/3#

#5/(x^2+x-6)=2+(x-3)/(x-2)#

#:.5/((x+3)(x-2))=2/1+(x-3)/(x-2)#

#:.(5=2(x+3)(x-2)+(x-3)(x+3))/((x+3)(x-2))#

multiply both sides by #(x+3)(x-2)#

#:.5=2(x+3)(x-2)+(x-3)(x+3)#

#:.5=2x^2+2x-12+x^2-9#

#:.5+12+9=3x^2+2x#

#:.3x^2+2x=26#

#:.3x^2+2x-26=0#

use quadratic rule:

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#a=3,b=2,c=-26#

#:.x=(-(2)+-sqrt((2)^2-4(3)(-26)))/(2(3))#

#:.x=(-2+-sqrt(316))/6#

#:.(-2+-sqrt(79*2*2))/6#

#:.(-2+-2sqrt(79))/6#

#:.x=(-2+2sqrt79)/6 or x=(-2-2sqrt79)/6#

#:.color(blue)(x=-1/3+(sqrt(79))/3 or x=-1/3-sqrt(79)/3#