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Standard XII

Mathematics

Higher Order Equations

If α , β ar...

Question

If $α, β$ are roots of the equation $a x^{2} + b x + c = 0$ , where $a, b, c$ are distinct real values, then $(1 + α + α^{2}) (1 + β + β^{2})$ is

1 - [(\frac{b + c - 2 c}{a}) - (\frac{b^{2} + c^{2} + b c}{a^{2}})]

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1 - [(\frac{b + c - 2 c}{a}) - (\frac{b^{2} + c^{2} + b c}{a^{3}})]

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1 - [(\frac{a + c - 2 c}{a}) - (\frac{a^{2} + c^{2} + b c}{a^{2}})]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 - [(\frac{a + b - 2 c}{a}) - (\frac{c^{2} + b^{2} + a c}{a^{2}})]

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Solution

The correct option is A $1 - [(\frac{b + c - 2 c}{a}) - (\frac{b^{2} + c^{2} + b c}{a^{2}})]$
$α + β = \frac{- b}{a}$

$α β = \frac{- c}{a}$

$(1 + α + α^{2}) (1 + β + β^{2})$

$= 1 + β + β^{2} + α + α β + α β^{2} + α^{2} + α^{2} β + α^{2} β^{2}$

$= 1 + (α + β) + (α^{2} + β^{2}) + α β + α^{2} β^{2} + α β (α + β)$

$= 1 + (α + β) + [(α + β)^{2} - 2 α β] + α β + α^{2} β^{2} + α β (α + β)$

$= 1 - \frac{b}{a} + \frac{b^{2}}{a^{2}} + 2 \frac{c}{a} - \frac{c}{a} + \frac{c^{2}}{a^{2}} - \frac{c}{a} (\frac{- b}{a})$

$= 1 - [(\frac{b + c - 2 c}{a}) - (\frac{b^{2} + c^{2} + b c}{a^{2}})]$

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Similar questions

Q. Lets consider quadratic equation

a x^{2} + b x + c = 0

where

a, b, c \in R

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α + β = - \frac{b}{a}, α β = \frac{c}{a}

and the equation can be written as

a x^{2} + b x + c = a (x - α) (x - β)

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a_{1}, a_{2}, a_{3}, a_{4}

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and if

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Q. If three real numbers

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If α,β are roots of the equation ax2+bx+c=0, where a,b,c are distinct real values, then (1+α+α2)(1+β+β2) is

If $α, β$ are roots of the equation $a x^{2} + b x + c = 0$ , where $a, b, c$ are distinct real values, then $(1 + α + α^{2}) (1 + β + β^{2})$ is