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# Extraneous solutions?

0
826
2 Jun 17, 2015

#1
+13

2 - 12/(4 - x)  = (2x + 4) / (x^2 - 16)   we can rewrite this as

2 + 12/(x - 4)   = 2(x + 2)/ [(x - 4) (x + 4) ]        multiply through by [(x - 4) (x + 4) ]

2 [(x - 4) (x + 4) ] + 12(x + 4)  = 2(x + 2)        divide thrrough by 2

[(x - 4) (x + 4)] + 6(x + 4)  = x + 2    simplify

x^2 - 16 + 6x + 24  =  x + 2

x^2 + 6x + 8   = x + 2

x^2 + 5x + 6    =  0      factor

(x + 3) (x + 2)  = 0      and setting each factor to 0 we have x = -2   or  x = -3

{There are no extraneous solutions }   Jun 17, 2015

#1
+13

2 - 12/(4 - x)  = (2x + 4) / (x^2 - 16)   we can rewrite this as

2 + 12/(x - 4)   = 2(x + 2)/ [(x - 4) (x + 4) ]        multiply through by [(x - 4) (x + 4) ]

2 [(x - 4) (x + 4) ] + 12(x + 4)  = 2(x + 2)        divide thrrough by 2

[(x - 4) (x + 4)] + 6(x + 4)  = x + 2    simplify

x^2 - 16 + 6x + 24  =  x + 2

x^2 + 6x + 8   = x + 2

x^2 + 5x + 6    =  0      factor

(x + 3) (x + 2)  = 0      and setting each factor to 0 we have x = -2   or  x = -3

{There are no extraneous solutions }   CPhill Jun 17, 2015
#2
+10

Thank you CPhill!

Jun 17, 2015