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# If $\alpha$ and $\beta$, $\alpha$ and $\gamma$, $\alpha$ and $\delta$ are the roots of the equations $a{x^2} + 2bx + c = 0$, $2b{x^2} + cx + a = 0$ and $c{x^2} + ax + 2b = 0$ respectively, where $a,b,c$ are positive real numbers, then $\alpha + {\alpha ^2} =$A. $- 1$B. $0$C. $abc$D. $a + 2b + c$E. $abc$ Verified
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Hint: Use the fact that $\alpha$ is a root for all the equations. Substitute $x = \alpha$ in all the equations and all three equations to get a single equation. Write the obtained equation as a product of two factors and use the fact that $a,b,c$ are positive real numbers.

Formula used: If $\phi$ is a root of the polynomial $f(x) = a{x^2} + bx + c$, then $f(\phi ) = a{\phi ^2} + b\phi + c = 0$

Complete step-by-step solution:
Let $f(x) = a{\alpha ^2} + 2b\alpha + c$, $g(x) = 2b{\alpha ^2} + c\alpha + a$ and $h(x) = c{\alpha ^2} + a\alpha + 2b$
Since $\alpha$ is a root for all the equations, $f(\alpha ) = 0$, $g(\alpha ) = 0$ and $h(\alpha ) = 0$
$a{\alpha ^2} + 2b\alpha + c = 0$
$2b{\alpha ^2} + c\alpha + a = 0$
$c{\alpha ^2} + a\alpha + 2b = 0$
Adding the three equations above we get,
$a{\alpha ^2} + 2b\alpha + c + 2b{\alpha ^2} + c\alpha + a + c{\alpha ^2} + a\alpha + 2b = 0$
$\left( {a + 2b + c} \right)\left( {{\alpha ^2} + \alpha + 1} \right) = 0$
$a + 2b + c \ne 0$. Therefore,
${\alpha ^2} + \alpha + 1 = 0$
${\alpha ^2} + \alpha = - 1$
Therefore, the correct answer is Option A. $- 1$.

Note: If three numbers are positive numbers then their sum must also be a positive number and therefore, they cannot be 0. That is why $a + 2b + c \ne 0$. Alternatively, we can say the same by writing three inequalities and adding them as follows: $a > 0$, $2b > 0$, $c > 0$. Adding the three inequalities we get $a + 2b + c > 0$.

Last updated date: 30th Sep 2023
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