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grangergirl
grangergirl grangergirl
  • 04-11-2014
  • Mathematics
Answered

\(0= \frac{3 x^{2}-18x+24 }{ x^{2} -2x-8}\)


Solve for x
Please show your work!!!

Thanks in advance :)

Verified Answer :

konrad509
konrad509 konrad509
  • 04-11-2014

\(D:x^2-2x-8 ot=0\\ D:x^2-4x+2x-8 ot=0\\ D:x(x-4)+2(x-4) ot=0\\ D:(x+2)(x-4) ot=0\\ D:x ot =-2 \wedge x ot =4\\\\ \frac{3x^2-18x+24}{x^2-2x-8}=0\\ \frac{3(x^2-6x+8)}{(x+2)(x-4)}=0\\ \frac{3(x^2-4x-2x+8)}{(x+2)(x-4)}=0\\ \frac{3(x(x-4)-2(x-4))}{(x+2)(x-4)}=0\\ \frac{3(x-2)(x-4)}{(x+2)(x-4)}=0\\ \frac{3(x-2)(x-4)}{(x+2)(x-4)}=0\\ \frac{3(x-2)}{(x+2)}=0|\cdot(x+2)\\ 3(x-2)=0|:3\\ x-2=0\\ \boxed{x=2}\)

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