Question

If $c,d$ are zeros of $x^2-10ax-11b$ and $a,b$ are zeros of $x^2-10cx-11d$, then value of $a+b+c+d$ is :

• A. $1210$
• B. $-1$
• C. $2530$
• D. $-11$

Answer 1

The correct option is A $1210$

From properties of roots,

$a+b=10c$ - (1), $c+d=10a$ - (2) , $a.b=-11d$ & $cd=-11b$

subtracting (2) from (1) $(x.c)+(b-d)=10(c-a)$

$\therefore b-d=11(c-a)$ - (3)

As a is root of $x^2-10cx-11d=0,a^2-10ac-11d=0$ - (4)

Similarly, $c^2-10ca-11b=0$ - (5)

subtracting (4) from (5), $c^2-a^2=11(b-d)$

$\therefore (c+a)(c-a)=11\times 11(c-a)$ (from 3)

$\therefore c+a=121$ Adding (1) & (2), $a+b+c+d=10\times 121$

$=1210$