Theory Of Equations Ques 43
- 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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Answer:
Correct Answer: 43.(1210)
Solution:
- Here,
$ a+b=10 c \text { and } c+d=10 a $
$\begin{aligned} \Rightarrow & & (a-c)+(b-d) & =10(c-a) \ \Rightarrow & & (b-d) & =11(c-a) ……(i)\end{aligned}$
Since, ’ $c$ ’ is the root of $x^{2}-10 a x-11 b=0$
$ c^{2}-10 a c-11 b=0 ……(ii) $
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 ……(iii) \end{aligned} $
On subtracting Eq. (iv) from Eq. (ii), we get
$\left(c^{2}-a^{2}\right) =11(b-d) …… (iv)$
$\therefore \quad(c+a)(c-a) =11 \times 11(c-a) [from Eq. (i)]$
$\Rightarrow c+a =121 $
$\therefore a+b+c+d =10 c+10 a $
$ =10(c+a)=1210$
BETA
