( $a, b$ ) roots of $x^2-10 c x-11 d=0$

( $c, \mathrm{~d}$ ) roots of $x^2-10 a x-11 b=0$

We know sum of roots

$$ \quad \begin{aligned} a+b=10 c \text { and } c+d & =10 a \\ \Rightarrow \quad a+b+c+d & =10(c+a) \end{aligned} $$

$$ \Rightarrow \quad(b+d)=9(a+c) $$

Now, we calculate $(c+a)$

Here product of root

$$ \begin{aligned} & & a b=-11 d \text { and } c d & =-11 b \\ \Rightarrow & & a b c d & =121 b d \\ \Rightarrow & & a c & =121 \end{aligned} $$

$$ \begin{aligned} & \text { Also } \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad a^2-10 c a-11 d=0 \\ &\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, c^2-10 a c-11 b=0 \\ & \Rightarrow \quad \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, c^2+a^2-20 a c-11(b+\mathrm{d})=0 \\ & \Rightarrow\,\,\, c^2+a^2+2 a c-22 a c-11 \times 9(a+c)=0 \\ & \Rightarrow \quad(a+c)^2-22 \times 121-11 \times 9(a+c)=0 \\ & \Rightarrow\,\,\,\,\,\,\,\,\, \quad(a+c)^2-99(a+c)-22 \times 121=0 \end{aligned} $$

$$ \begin{aligned} a+c & =\frac{99 \pm \sqrt{9801+88 \times 121}}{2} \\ & =\frac{99 \pm \sqrt{9801+10648}}{2} \\ & =\frac{99 \pm \sqrt{20449}}{2}=\frac{99 \pm 143}{2} \\ & =\frac{242}{2}, \frac{-44}{2}=121,-22 \end{aligned} $$

Discard the value of $a+c=-22$

$$ \begin{aligned}Hence,\,\, a+b+c+d & =121 \times 10 \\ & =1210 \end{aligned} $$

JEE Advanced Subjects