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Consider the equation $x^n-5 x^{n-1}+6 x^{n-2}=0$, where $n$ is a positive integer. What is the sum of all possible real roots of this equation?
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The correct answer is:
$5$
To find the sum of all possible real roots of the equation, we can factor out $x^{n-2}$ from the equation:$x^{n-2}\left(x^2-5 x+6\right)=0$
Now, we solve $x^2-5 x+6=0$ for real roots. Factoring this quadratic equation, we get:$(x-2)(x-3)=0$
So, the roots of $x^2-5 x+6=0$ are $x=2$ and $x=3$.
Since the original equation is $x^n-5 x^{n-1}+6 x^{n-2}=0$, the sum of all possible real roots is the sum of the roots of $x^2-5 x+6=0$, which is $2+3=5$.
Therefore, the correct answer is 1) 5 .
Now, we solve $x^2-5 x+6=0$ for real roots. Factoring this quadratic equation, we get:$(x-2)(x-3)=0$
So, the roots of $x^2-5 x+6=0$ are $x=2$ and $x=3$.
Since the original equation is $x^n-5 x^{n-1}+6 x^{n-2}=0$, the sum of all possible real roots is the sum of the roots of $x^2-5 x+6=0$, which is $2+3=5$.
Therefore, the correct answer is 1) 5 .
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