Find the roots - - - are the roots of ax2-bx-c-0 then 1- -22- -2-

The correct option is A 1a^2(a^2+b^2+c^2+ab+bc+ca) Now, [ a x^ 2 +bx+c=0; α +β = b / a ,αβ = c ...

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Standard XII

Mathematics

Integration of Piecewise Continuous Functions

Find the root...

Question

Find the roots $α, β$ are the roots of $a x^{2} + b x + c = 0$ then $(1 + α + α^{2}) (2 + β + β^{2}) =$

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

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

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

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

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Solution

The correct option is A $\frac{1}{a^{2}} (a^{2} + b^{2} + c^{2} + a b + b c + c a)$
Now,
$\begin{matrix} a x^{2} + b x + c = 0 α + β = - \frac{b}{a}, α β = \frac{c}{a} {(α - β)}^{2} = {(α + β)}^{2} - 4 α β = \frac{b^{2}}{a^{2}} - \frac{4 c}{a} = \frac{b^{2} - 4 a c}{a^{2}} α - β = \frac{\sqrt{b^{2} - 4 a c}}{a} ∴ 2 α = \frac{- b + \sqrt{b^{2} - 4 a c}}{a} α = \frac{- b \sqrt{b^{2} - 4 a c}}{2 a} \end{matrix}$
$\begin{matrix} ∴ β = - \frac{b}{a} - α = - \frac{b}{a} - (\frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}) = \frac{2 b}{2 a} + \frac{b}{2 a} - \frac{\sqrt{b^{2} - 4 a c}}{2 a} = \frac{- b}{2 a} - \frac{\sqrt{b^{2} - 4 a c}}{2 a} β == \frac{- b}{2 a} - \frac{\sqrt{b^{2} - 4 a c}}{2 a} \end{matrix}$

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Similar questions

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Find the roots α, β are the roots of ax2+bx+c=0 then (1+α+α2)(2+β+β2)=

Find the roots $α, β$ are the roots of $a x^{2} + b x + c = 0$ then $(1 + α + α^{2}) (2 + β + β^{2}) =$