Quadratic EquationHard
Question
Let $a,b,c \in R,a \neq 0$ such that a and $4a + 3\text{ }b + 2c$ have the same sign. Then the equation $ax^{2} + bx + c = 0$ must have
Options
A.both roots in $(1,2)$
B.no roots in $(1,2)$
C.not both roots in $(1,2)$
D.exactly one root in $(1,2)$
Solution
Let $\alpha,\beta$ be roots of the given equation $\frac{4a + 3b + 2c}{2} > 0 \Rightarrow 4 - 3(\alpha + \beta) + 2\alpha\beta > 0$
$$\Rightarrow \ \begin{matrix} & \alpha\beta - 2\alpha - \beta + 2 + \alpha\beta - \alpha - 2\beta + 2 > 0 \\ \Rightarrow & \ (\alpha - 1)(\beta - 2) + (\alpha - 2)(\beta - 1) > 0 \end{matrix}$$
If $\alpha,\beta$ both belong to $(1,2)$
$$\begin{matrix} \Rightarrow & (\alpha - 1)(\beta - 2) < 0\text{~and~}(\alpha - 2)(\beta - 1) < 0 \\ \Rightarrow & \frac{4a + 3b + 2c}{a} < 0 \end{matrix}$$
which is contradiction
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