Bored with Vieta's formulae?

Algebra Level 4

If α , β \alpha,\beta are the roots of the equation x 2 − p ( x + 1 ) − c = 0 x^{2}-p(x+1)-c =0 (where p , c p,c are real numbers ,not equal to zero) then α 2 + 2 α + 1 α 2 + 2 α + c + β 2 + 2 β + 1 β 2 + 2 β + c \large \frac{\alpha^{2}+2\alpha+1}{\alpha^{2}+2\alpha+c} +\frac{\beta^{2}+2\beta+1}{\beta^{2}+2\beta+c} is equal to:


The answer is 1.

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Keshav Tiwari by Keshav Tiwari , 23, India

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1 solution

Manish Dash
Sep 1, 2015

F r o m V i e t a ′ s R u l e , α + β = p α β = − ( p + c ) T h e g i v e n e q u a t i o n h a s α a n d β a s i t s r o o t s . H e n c e α 2 − p ( α + 1 ) = c − − − − − − − − − − ( i ) β 2 − p ( β + 1 ) = c − − − − − − − − − − ( i i ) α 2 + 2 α + 1 α 2 + 2 α + c + β 2 + 2 β + 1 β 2 + 2 β + c = α 2 + 2 α + 1 α 2 + 2 α + α 2 − p α − p + β 2 + 2 β + 1 β 2 + 2 β + β 2 − p β − p = ( α + 1 ) 2 2 α ( α + 1 ) − ( α + β ) ( α + 1 ) + ( β + 1 ) 2 2 β ( β + 1 ) − ( β + 1 ) ( α + β ) = ( α + 1 ) 2 ( α + 1 ) ( α − β ) + ( β + 1 ) 2 ( β + 1 ) ( β − α ) = ( α + 1 ) ( α − β ) + ( β + 1 ) ( β − α ) = ( α + 1 ) ( α − β ) − ( β + 1 ) ( α − β ) = α − β α − β = 1 From\quad Vieta's\quad Rule,\\ \alpha +\beta \quad =\quad p\\ \alpha \beta \quad =\quad -(p+c)\\ \\ The\quad given\quad equation\quad has\quad \alpha \quad and\quad \beta \quad as\quad its\quad roots.\quad Hence\quad \\ { \alpha }^{ 2 }-p(\alpha +1)\quad =\quad c\quad ----------(i)\\ { \beta }^{ 2 }-p(\beta +1)\quad =\quad c\quad ----------(ii)\\ \\ \frac { { \alpha }^{ 2 }+2\alpha +1 }{ { \alpha }^{ 2 }+2\alpha +c } +\frac { { \beta }^{ 2 }+2\beta +1 }{ { \beta }^{ 2 }+2\beta +c } \quad \\ =\quad \frac { { \alpha }^{ 2 }+2\alpha +1 }{ { \alpha }^{ 2 }+2\alpha +{ \alpha }^{ 2 }-p\alpha -p } +\frac { { \beta }^{ 2 }+2\beta +1 }{ { \beta }^{ 2 }+2\beta +{ \beta }^{ 2 }-p\beta -p } \\ =\frac { (\alpha +1{ ) }^{ 2 } }{ 2\alpha (\alpha +1)-(\alpha +\beta )(\alpha +1) } +\frac { (\beta +1{ ) }^{ 2 } }{ 2\beta (\beta +1)-(\beta +1)(\alpha +\beta ) } \\ =\quad \frac { (\alpha +1{ ) }^{ 2 } }{ (\alpha +1)(\alpha -\beta ) } +\frac { (\beta +1{ ) }^{ 2 } }{ (\beta +1)(\beta -\alpha ) } \\ =\quad \frac { (\alpha +1{ ) } }{ (\alpha -\beta ) } +\frac { (\beta +1{ ) } }{ (\beta -\alpha ) } \\ =\quad \frac { (\alpha +1{ ) } }{ (\alpha -\beta ) } -\frac { (\beta +1{ ) } }{ (\alpha -\beta ) } \\ =\quad \frac { \alpha -\beta }{ \alpha -\beta } \\ =\quad \boxed{1}

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Moderator note:

Well, we also have to ensure that the denominator is non-zero right?

Well done ! : ) :)

Keshav Tiwari - 5 years, 9 months ago

Can you please explain me i , i i i,ii

Department 8 - 5 years, 9 months ago

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As α \alpha and β \beta are the roots of the given equation , when put in place of x , they make the equation zero.

Keshav Tiwari - 5 years, 9 months ago

you could also add the first two equations and get α + β + α β = c a n d s u b s t i t u t e i t . \alpha +\beta +\alpha \beta =c\\ and\quad substitute\quad it.

Rishi Sharma - 5 years, 9 months ago

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