JEE Advanced · 2006 · 09 Apr 2006 · Quadratic Equations and Inequalities

From Vieta's formulas:

• $a+b=10c$ and $ab=-11d$
• $c+d=10a$ and $cd=-11b$

From the product relations: $ab\cdot cd=(-11d)(-11b)=121bd$, so $ac=121$.

Adding the sum relations: $(a+b)+(c+d)=10c+10a$, which gives $(b+d)=9(a+c)$.

Since a and c are roots of their respective equations:

• $a^2-10ac-11d=0$
• $c^2-10ac-11b=0$

Adding: $a^2+c^2-20ac-11(b+d)=0$

Substituting $ac=121$ and $b+d=9(a+c)$:



Solving: $(a+c)=121$ or $-22$ (rejected as it gives negative values).

Therefore: $a+b+c+d=(a+c)+(b+d)=121+9(121)=121+1089=1210$.