Theory of Equations 1 Question 44
45. If $x^{2}-10 a x-11 b=0$ have roots c and d. $x^{2}-10 c x-11 d=0$ have roots $a$ and $b$, then find $a+b+c+d$.
$(2006,6 M)$
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Solution:
- Here,
$$ a+b=10 c \text { and } c+d=10 a $$
$\begin{aligned} \Rightarrow & & (a-c)+(b-d) & =10(c-a) \ \Rightarrow & & (b-d) & =11(c-a)\end{aligned}$
Since, ’ $c$ ’ is the root of $x^{2}-10 a x-11 b=0$
$$ c^{2}-10 a c-11 b=0 $$
Similarly, ’ $a$ ’ is the root of
$$ \begin{aligned} x^{2}-10 c x-11 d & =0 \\ \Rightarrow \quad a^{2}-10 c a-11 d & =0 \end{aligned} $$
On subtracting Eq. (iv) from Eq. (ii), we get
$$ \begin{array}{rlrl} \left(c^{2}-a^{2}\right) & =11(b-d) \\ & \therefore \quad(c+a)(c-a) & =11 \times 11(c-a) \\ \Rightarrow & & c+a & =121 \\ \therefore & & a+b+c+d & =10 c+10 a \\ & =10(c+a)=1210 \end{array} $$
