If the roots of the equation $$ax^2 + bx + c = 0$$ are of the form $$\frac{\alpha}{(\alpha - 1)}$$ and $$\frac{\alpha + 1}{\alpha}$$, then the value of $$(a+b+c)^2$$ is 

(a) $$2b^2 - ac$$    

(b) $$b^2 - 2ac$$

(c) $$b^2 - 4ac$$

(d) $$4b^2 - 2ac$$

Please let me know the complete solution.

Since the roots of the given equation are of the form $$\frac{\alpha}{\alpha-1}$$ and $$\frac{\alpha+1}{\alpha}$$

Therefore , 

Product of the Roots = $$\frac{\alpha}{\alpha-1} \cdot  \frac{\alpha+1}{\alpha} = \frac{\alpha+1}{\alpha-1} =  \frac{c}{a}$$ 

Sum of Roots = $$\frac{\alpha}{\alpha-1} + \frac{\alpha+1}{\alpha} = \frac{2\alpha^2-1}{\alpha(\alpha-1)} = -\frac{b}{a} \cdots (1) $$

By Applying Componendo and Dividendo

$$\implies \alpha = \frac{c+a}{c-a} \cdots (2) $$ 

Plugging $$(2)$$ in $$(1)$$

$$\implies \frac{2\left(\frac{c+a}{c-a}\right)^2-1}{\left(\frac{c+a}{c-a}\right)\left(\frac{c+a}{c-a}-1\right)} = -\frac{b}{a} $$ 

Solving the above equation

$$\implies\frac{2\left ( c+a \right )^2-\left ( c-a \right )^2}{\left ( c+a \right )\cdot\left ( (c+a)-(c-a) \right )} = -\frac{b}{a}$$ 

$$\implies\frac{c^2+a^2+6ac}{c+a}=-2b$$

$$\implies c^2+a^2+2ab+2bc+6ca = 0 \cdots (3)$$

Evaluating the given expression

$$\left(a+b+c\right)^2=\left(a^2+b^2+c^2+2ab+2bc+2ca\right)$$

$$\left(a+b+c\right)^2=\left(a^2+c^2+2ab+2bc+6ca\right)+\left(b^2-4ac\right) \cdots(4)$$

Plugging $$(3)$$ in $$(4)$$

$$\left(a+b+c\right)^2=\left(b^2-4ac\right)$$

Hence , Correct Option is (C)