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Can someone help me understand the solutions to this homogeneous equation?
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\[\alpha ^{n}-5\alpha ^{n-1}+6\alpha^{n-2}=0\]
The solutions for alpha is 2 and 3, how come?
\[a^n-5a^{n-1}+6a^{n-2}=0\] \[a^n-5a^{n-1}+6a^{n-2}=a^{n-2}(a^2-5a+6)=0\] \[a^n \neq 0\] that means \[a^2-5a+6=0\] solve for a
Oh nice! that make sense but why is it not negative though?
because if you actually factor the quadratic equation and solve for a, the answers will be positive...
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or because for negative numbers, even and odd roots will have different signs making it much more difficult for an actual solution for a, if any
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