Problem of relations between the coefficients of a quadratic equation.how to proceed?

Question

I have tried several ways to solve this problem of using the root properties of a second degree equation but I can't come up with anything favorable.

If the roots of $ax^{2}+bx+c=0$ are of the form $\dfrac{\lambda }{\lambda -1}$ and $\dfrac{\lambda+1 }{\lambda}$, then the value $(a+b+c)^{2}$ in terms of $a$, $b$, $c$ is?

Options

$(a) \quad b^{2}-4ac$

$(b) \quad b^{2}-2ac$

$(c) \quad 2b^{2}-ac$

$(d) \quad 4b^{2}-2ac$

$(e) \quad b^{2}-4abc$

Answer 1

\begin{align} \frac{c}{a} &= \frac{\lambda}{\lambda-1} \times \frac{\lambda+1}{\lambda} \\ &= \frac{\lambda+1}{\lambda-1} \\ \lambda &= \frac{c+a}{c-a} \\ x_1 &= \frac{\lambda}{\lambda-1} \\ &= \frac{c+a}{2a} \\ x_2 &= \frac{\lambda+1}{\lambda} \\ &= \frac{2c}{c+a} \\ -\frac{b}{a} &= x_1+x_2 \\ &= \frac{(c+a)^2+4ac}{2a(c+a)} \\ -2b(c+a) &= (c+a)^2+4ac \\ 0 &= (c+a)^2+2b(c+a)+4ac \\ \color{red}{b^2}-4ac &= (c+a)^2+2b(c+a) \color{red}{+b^2} \\ (a+b+c)^2 &= b^2-4ac \end{align}

Answer 2

I'll suggest you hints how to do it in a way-

1. Product of roots is c/a. Use componendo-dividendo to express lambda in terms of c and a.
2. Sum of roots is -b/a. Substitute lambda in this from above to get an equation in a, b and c.
3. Simplify this equation in the form $(a+b+c)^2=..... $

Answer 3

Define $f(x)=ax^2+bx+c$ then $$ f(x)=ax^2+bx+c=a\left(x-\frac{\lambda}{\lambda-1}\right)\left(x-\frac{\lambda+1}{\lambda}\right) $$ Hence $$ (a+b+c)^2=(f(1))^2=a^2 \left(1 - \frac{\lambda}{\lambda-1}\right)^2 \left(1 - \frac{\lambda+1}{\lambda}\right)^2\\ = a^2 \frac{1}{(\lambda-1)^2} \frac{1}{\lambda^2} = a^2 \left(\frac{\lambda}{\lambda-1} - \frac{\lambda+1}{\lambda}\right)^2 \\= a^2 (r-s)^2 = a^2 ((r+s)^2 - 4rs)\\=a^2(-b/a)^2 - 4 a^2 (c/a)=b^2-4ac $$ where $r=\lambda/(\lambda-1)$ and $s=(\lambda+1)/\lambda$