What are the zeros of the polynomial equation f(x)=x^2-12x+20?

1 answer:

8 0

$f(x)=x^2-12x+20\\\\\text{The zeros:}\\\\f(x)=0\to x^2-12x+20=0\\\\x^2-10x-2x+20=0\\\\x(x-10)-2(x-10)=0\\\\(x-10)(x-2)=0\iff x-10=0\ \vee\ x-2=0\\\\\boxed{x=10\ \vee\ x=2}$

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$\dfrac{1}{1\times 3}+\dfrac{1}{3\times 5}=\dfrac{5+1}{1\times 3\times 5}=\dfrac{2}{5}\quad\text{sum of 2 terms}\\\\\dfrac{2}{5}+\dfrac{1}{5\times 7}=\dfrac{14+1}{5\times7}=\dfrac{3}{7}\quad\text{sum of 3 terms}$

Suppose we say the sum of n terms is (n/(2n+1)), the next term in the series will be 1/((2n+1)(2n+3)) and adding that to the presumed sum gives ...

$\dfrac{n}{2n+1}+\dfrac{1}{(2n+1)(2n+3)}=\dfrac{n(2n+3)+1}{(2n+1)(2n+3)}=\dfrac{2n^2+3n+1}{(2n+1)(2n+3)}\\\\=\dfrac{(2n+1)(n+1)}{(2n+1)(2n+3)}=\dfrac{n+1}{2n+3}\text{ matches }\dfrac{(n+1)}{2(n+1)+1}$