3x^(6)e^(-2x)-9x^(7)e^(-2x)=0 has two roots Where do I even start?
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I would start by dividing through by any non-zero numbers that are common to both sides of the equation. (I presume you know that $\displaystyle e^x$ is not 0 for ny x.)
Originally Posted by RD33 3x^(6)e^(-2x)-9x^(7)e^(-2x)=0 has two roots Where do I even start? Factor it: $\displaystyle 3x^6e^{-2x}(1-3x)=0~.$
Ok I got the answer 0 and 1/3 but how did you factor it. Where does the (1-3x) come from?
Originally Posted by RD33 Ok I got the answer 0 and 1/3 but how did you factor it. Where does the (1-3x) come from? Can you factor $\displaystyle 3a^6b-9a^7b~?$ What do you get?
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