anonymous
  • anonymous
3/x+2 +2/x-2=8/(x+2)(x-2) Solve and tell if its extraneous
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
anonymous
  • anonymous
now you can have extraneous solutions because you have denominators
anonymous
  • anonymous
:)

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anonymous
  • anonymous
\[\frac{x}{x+2}+\frac{2}{x-2}=\frac{8}{(x+2)(x-2)}\] you have a choice, you can either add on the left, or multiply both sides by \((x+2)(x-2)\) to clear the fractions
anonymous
  • anonymous
multiply
anonymous
  • anonymous
ok then the \((x+2)(x-2)\) cancels on the right leaving \(8\)
anonymous
  • anonymous
on the left the first term will be \(x(x-2)\) and the second term will be \(2(x+2)\)
anonymous
  • anonymous
by cancelling that leaves you with the equation \[x(x-2)+2(x+2)=8\] multiply out and solve the resulting quadratic
anonymous
  • anonymous
x^2-2x+2x+4=8 x^2+4=8 x^2=4
anonymous
  • anonymous
ok and if \(x^2=4\) then \(x=?\)
anonymous
  • anonymous
x=+ or - 2
anonymous
  • anonymous
right
anonymous
  • anonymous
and in this case
anonymous
  • anonymous
BOTH are extraneous, because both values would make the denominator equal to zero
anonymous
  • anonymous
:) Thanks!
anonymous
  • anonymous
yw

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