Question

If alpha and beta are the zeroes of the polynomial p(x) = x2 + x + 1, then find the sum of their reciporcal​

Answer 1

${ \bold{ \green{ \underline{ \underline{ANSWER}} : - }}}$

${ \bold{ \underline{Given \: \: function}} : - } \\ \\ { \bold{ \pink{p(x) = {x}^{2} + x + 1}}} \\ \\ { \bold{ \underline{To \: \: find} : - }} \\ \\ { \bold{ \red{ = \frac{1}{ \alpha } + \frac{1}{ \beta } }}} \\ \\ { \bold{ \red{ = \frac{ \alpha + \beta }{ \alpha \beta } }}} \\ \\ { \bold{ \red{ = \frac{ \frac{ - b}{a} }{ \frac{c}{a} } }}} \\ \\ { \bold{ \red{ = - \frac{b}{c} }}} \\ \\ { \bold{ \red{ = - 1}}} \\ \\ \\ \\ { \bold{ \boxed{{ \huge{\red{FOLLOW \: \: ME... }}}}}}$

Answer 2

Answer:   -1

Step-by-step explanation:

                                                p(x) = x^2 + x + 1

 zeroes are the values which when we put in place of x  makes p(x)=0.

 Therefore, the values at which x^2 + x + 1 =0 are

 1. ( -1 + i√3 ) / 2  which we call omega(w)

 2. ( -1 - i√3 ) / 2 which we call omega²(w²)

Therefore sum of their reciprocals is   =    1/w  +  1/w²

                                                                =   ( w + w²) / w³

                                                                 =   -1 / 1

                                                                 =   -1 ( ANSWER)

     AS    1.   1 + w + w² = 0

              2.  w³ = 1

hope you will understand it !

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