Question

if alpha and beta are the zeroes of the polynomial p(x)=x²+x+1 then find the sum of their reciprocals 1/alpha+1/beta

Answer 1

hope it helps thou..........

Answer 2

Answer:

$Sum \: of \: reciprocals\\of \: the \: zeroes\\=\frac{1}{\alpha}+\frac{1}{\beta}=1$

Step-by-step explanation:

$Given \: \alpha \: and \:\beta\\ are \: the \: zeroes \:of\\the\: polynomial \:p(x)=x^{2}+x+1$

$Compare \: p(x) \: with \\ax^{2}+bx+c,\:we\:get$

$a=1,b=1\:and \:c=1$

$Now,\\i) Sum \:of \: the \: zeroes = \frac{-b}{a}$

$\implies \alpha+\beta\\= \frac{-1}{1}\\=-1---(1)$

$ii) Product \: of \: the \: zeroes = \frac{c}{a}$

$\implies \alpha\beta\\= \frac{1}{1}\\=1---(2)$

$Sum \: of \: reciprocals\\of \: the \: zeroes\\=\frac{1}{\alpha}+\frac{1}{\beta}\\=\frac{\alpha+\beta}{\alpha\beta}\\=\frac{-1}{1}\\=-1$

Therefore,

$Sum \: of \: reciprocals\\of \: the \: zeroes\\=\frac{1}{\alpha}+\frac{1}{\beta}=1$

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