Problem :
If the roots of $ax^2+bx+c=0$ are of the form $\frac{m}{m-1}$ and $\frac{m+1}{m}$ then find the value of $(a+b+c)^2$
My approach :
Let $\alpha, \beta$ are the two roots of the given equation. then $\frac{1}{\alpha}+\beta =2$
Also we know that sum of the roots $= \frac{-b}{a}$ and product of the roots $\frac{c}{a}$ but how to find $(a+b+c)^2$ not getting any idea further please help thanks.
Say the roots are $\frac{m}{m-1}$ and $\frac{m+1}{m}$.
Then we have that, $$\frac{m}{m-1}+\frac{m+1}{m}=-\frac{b}{a}$$ $$\frac{m}{m-1}\cdot \frac{m+1}{m}=\frac{c}{a}$$
So we can conclude that $$\frac{2m^2-1}{m(m-1)}=-\frac{b}{a}$$ $$\frac{m+1}{m-1}=\frac{c}{a}$$
So we can write that $$(a+b+c)^2=a^2(1+\frac{b}{a}+\frac{c}{a})^2$$ $$=a^2\left[1-\frac{2m^2-1}{m(m-1)}+\frac{m+1}{m-1}\right]^2$$ $$=a^2\left[\frac{m^2-m-2m^2+1+m^2+m}{m(m-1)}\right]^2$$ $$=\left[\frac{a}{m(m-1)}\right]^2$$
Now, again we can write that $$(x-\frac{m}{m-1})(x-\frac{m+1}{m})=0$$ $$x^2-(\frac{m}{m-1}+\frac{m+1}{m})x-\frac{m+1}{m-1}=0$$ $$m(m-1)x^2-(2m^2-1)x-(m^2+m)=0$$
By comparison with $ax^2+bx+c=0$, we can say that $a=rm(m-1)$.
So we have that $$(a+b+c)^2=r^2$$
See if you can progress any far with this result.