Question

if the zeros of the polynomial x^3 -3x^2 + x +1 are α-β , α and α+β , find α+β

Answer 1

Zeros of polynomial are alpha - beta , alpha , alpha + beta 

Sum of zeros =  (alpha - beta) + (alpha) + (alpha + beta) =  - b / a
                                       
3 * alpha    =   3 / 1 
                                               
alpha  =  1

product of zeros = (alpha - beta) *(alpha) *(alpha + beta)  =  - d /a

Now put value of alpha = 1
                   
(1 - beta) * 1 * (1+ beta)  =  -1
                                             
12 -  beta2   =  -1
                                             
- beta2 =  -1-1
                                             
beta 2= 2
                                               
beta =  + root 2

So alpha = 1   ,    beta = + root 2

alpha + beta = 1 + root 2

Thanks

Answer 2

☆HEY FRIEND!!! ☆

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HERE IS YOUR ANSWER ☞
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$given \: = > \\ \\ {x}^{3} - 3 {x}^{2} + x + 1 \\ \\ and \: also \: given \\ \\ the \: zeroes \: of \: the \: polynomial \: are \: \\ = > ( \alpha + \beta ) \: \: \alpha \: \: and \: \: \: ( \alpha - \beta ) \\ \\ we \: know \: that = > \\ \\ sum \: of \: the \: zeroes \: = - \frac{ {x}^{2} \: coefficient}{ {x}^{3} \: coefficient } \\ \\ ( \alpha - \beta ) + \alpha + ( \alpha + \beta ) = -\frac{ - 3}{1} \\ \\ = > 3 \alpha = 3 \\ \\ = > \alpha = 1 \: \: \: \: \: ......eq _{1} \\ \\ again \\ \\ we \: know \: that \\ \\ multiple \: of \: zeroes \: = - \frac{constant}{ {x}^{3} \: coefficient } \\ \\ = > ( \alpha - \beta ) \times ( \alpha ) \times ( \alpha + \beta ) = - \frac{1}{1} \\ \\ = > ( { \alpha }^{2} - { \beta }^{2} )( \alpha ) = - 1 \\ \\ plug \: the \: value \: of \: \alpha \: we \: get \\ \\ = > ( { 1}^{2} - { \beta }^{2} )( 1) = - 1 \\ \\ = > 1 - { \beta }^{2} = -1 \\ \\ = > { \beta }^{2} = 2 \\ \\ = > \beta = root2 \: \: \: \: \: ........eq _{2} \\ \\ now \\ \\ by \: addition \: of \: eq _{1} \: and \: eq _{2} \: \: \: we \: get \\ \\ = > \alpha + \beta \: = 1 + root2 \: \: \: \: \: \: ....answer$

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HOPE IT WILL HELP YOU ☺☺☺
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DEVIL_KING ▄︻̷̿┻̿═━一