If α, β and gamma are roots of equation {x^3} - 3{x^2} + x + 5 = 0 then y = Σ {α ^2} + α β gam

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4-2.Quadratic Equations and Inequations

hard

If $\alpha, \beta $ and $\gamma$ are the roots of equation ${x^3} - 3{x^2} + x + 5 = 0$ then $y = \sum {\alpha ^2} + \alpha \beta \gamma $ satisfies the equation

${y^3} + y + 2 = 0$

${y^3} - {y^2} - y - 2 = 0$

${y^3} + 3{y^2} - y - 3 = 0$

${y^3} + 4{y^2} + 5y + 20 = 0$

Solution

(b) Given equation ${x^3} – 3{x^2} + x + 5 = 0$.

Then $\alpha + \beta + \gamma = 3$, $\alpha \beta + \beta \gamma + \gamma \alpha = 1$, $\alpha \beta \gamma = – 5$

$y = \Sigma {\alpha ^2} + \alpha \beta \gamma = {(\alpha + \beta + \gamma )^2} – 2\,(\alpha \beta + \beta \gamma + \gamma \alpha ) + \alpha \beta \gamma $

= $9 – 2 – 5 = 2$

$\therefore $ $y = 2$

It satisfies the equation ${y^3} – {y^2} – y – 2 = 0$.

Standard 11

Mathematics

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normal

If $\alpha, \beta $ and $\gamma$ are the roots of equation ${x^3} - 3{x^2} + x + 5 = 0$ then $y = \sum {\alpha ^2} + \alpha \beta \gamma $ satisfies the equation

Solution

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Let $a$ ,$b$, $c$ , $d$ , $e$ be five numbers satisfying the system of equations

$2a + b + c + d + e = 6$
$a + 2b + c + d + e = 12$
$a + b + 2c + d + e = 24$
$a + b + c + 2d + e = 48$
$a + b + c + d + 2e = 96$ ,

then $|c|$ is equal to