Question $\displaystyle \frac{3x^2+4}{x^3+8}=\frac{3}{x+2} $ Attempt: $\displaystyle 3x^3+6x^2+4x+8=3x^3+24 $ Final answer I got is 2 Please check my answer because i think it is wrong Thank you
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Originally Posted by mj.alawami Question $\displaystyle \frac{3x^2+4}{x^3+8}=\frac{3}{x+2} $ Attempt: $\displaystyle 3x^3+6x^2+4x+8=3x^3+24 $ Final answer I got is 2 Please check my answer because i think it is wrong Thank you $\displaystyle 3x^3+6x^2+4x+8=3x^3+24 $ $\displaystyle 6x^2 +4x-16=0$ $\displaystyle 3x^2 + 2x-8=0 $ $\displaystyle 3x^2 + 6x-4x-8=0$ $\displaystyle 3x(x+2)-4(x+2)=0$ $\displaystyle (3x-4)(x+2)=0$ x=4/3 or x=(-2) But -2 is not allowed as denominator can't be 0
Last edited by ADARSH; Mar 9th 2009 at 08:07 AM. Reason: left last line
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