Mathematics - Quadratic Equation Question with Solution | TestHub

$\begin{array}{rl}& a(b-c)x^2+b(c-a)x+c(a-b)=0\\ & x=1\text{ is root }\therefore \text{ other root is }1\\ & \alpha +\beta =-\frac{b(c-a)}{a(b-c)}=2\\ & \Rightarrow -\mathrm{bc}+\mathrm{ab}=2\mathrm{ab}-2\mathrm{ac}\\ & \Rightarrow 2\mathrm{ac}=\mathrm{ab}+\mathrm{bc}\\ & \Rightarrow 2\mathrm{ac}=\mathrm{b}(\mathrm{a}+\mathrm{c})\\ & \Rightarrow 2\mathrm{ac}=15\mathrm{b}\dots (1)\\ & \Rightarrow 2\mathrm{ac}=15\left(\frac{36}{5}\right)=108\\ & \Rightarrow \mathrm{ac}=54\\ & \mathrm{a}+\mathrm{c}=15\\ & {\mathrm{a}}^2+{\mathrm{c}}^2+2\mathrm{ac}=225\\ & {\mathrm{a}}^2+{\mathrm{c}}^2=225-108=117\end{array}$