Mathematics - Quadratic Equation Question with Solution | TestHub

Here, $a+b=10c$ $$c+d=10a$$ On subtracting Eq. (ii) from Eq. (i), we get, $(a-c)+(b-d)=10(c-a)$ $\Rightarrow (b-d)=11(c-a)$ Since, $c$ is the root of $x^2-10ax+11b=0$ $$\Rightarrow c^2-10ac-11b=0$$ Similarly, $a$ is the root of $$\begin{array}{rl}\Rightarrow & x^2-10cx+11d=0\\ & a^2-10ca-11d=0\end{array}$$ On subtracting Eq. (v) from Eq. (iv), we get $$\begin{array}{rl}& \left(c^2-a^2\right)=11(b-d)\\ & \therefore (c+a)(c-a)=11\times 11(c-a)\{\text{ from (i) \& (ii) }\\ & \Rightarrow c+a=121.\\ & \therefore a+b+c+d=10c+10a\\ & =10(c+a)=1210\\ & \end{array}$$