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:

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

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

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

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

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

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

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

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

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