Repeated roots!

Let the other root be $\beta$ . Then by Vieta's formula , we have:

$\begin{cases} 2\alpha + \beta = - 3a & ...(1) \\ \alpha^2 + 2\alpha \beta = 3b & ...(2) \\ \alpha^2\beta = -c & ...(3) \end{cases}$

From $2\alpha \times (1) - (2)$ :

$\begin{aligned} 3\alpha^2 & = - 6a\alpha - 3b \\ \implies \alpha^2 + 2a \alpha + b & = 0 & ...(4) \end{aligned}$

From $(4)$ we note that $\alpha$ is a root of $x^2 + 2ax + b=0$ .

From $\alpha \times (2)$ :

$\begin{aligned} \alpha^3 + 2 \blue{\alpha^2 \beta} & = 3b\alpha & \small \blue{\text{Note that }(3): \ \alpha^2\beta = -c} \\ \implies \alpha^3 & = 3b\alpha + 2\blue c & ...(2a) \end{aligned}$

From $\alpha \times (4)$ :

$\begin{aligned} \blue{\alpha^3} + 2a \red{\alpha^2} + b\alpha & = 0 & \small \blue{\text{Note that }(2a): \ \alpha^3 = 3b\alpha + 2 c} \\ \blue{3b\alpha + 2 c} + 2a \red{(-2a \alpha - b)} + b\alpha & = 0 & \small \red{\text{From }(4): \ \alpha^3 = -2a \alpha - b} \\ 4b\alpha + 2c - 4a^2\alpha - 2ab & = 0 \\ 2(b-a^2)\alpha + c - ab & = 0 \\ \implies \alpha & = \frac {c-ab}{2(a^2-b)} \end{aligned}$

Therefore, the incorrect option is $\boxed{\alpha = \frac {ab-c}{a^2-b}}$ .

Note :- FTA means the Fundamental Theorem of Algebra view wiki - Fundamental Theorem of Algebra