# 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. $abc$

Last updated date: 21st Mar 2023

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Answer

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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 $a + 2b + c > 0$.

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