Let `aa n db` be the roots of the equation `x^2-10 c x-11 d=0` and those of `x^2-10 a x-11 b=0a r ec ,d then find the value of `a+b+c+d when a!=b!=c!=d

Answer

1210

Explanation

Roots of `x^(3)-10cx-11d=0` are a and b.
So, `a+b=10candab=-11d`
c and d are the roots of `x^(2)-10ax-11b=0`. So, `c+d=10aandcd=-11b`
From these reslts, we get
`a+b+c+d=10(a+c)`
`impliesb+d=9(a+c)` and `abcd=121bd`
`impliesac=121`
Also, we have `a^(2)-10ac-11d=0andc^(2)-10ac-11b=0`
`impliesa^(2)+c^(2)-20ac-11(b+d)=0`
`implies(a+c)^(2)-22ac-99(a+c)=0`
`implies(a+c)^(2)-99(a+c)-22xx121=0`
`implies121-99(a+c)=0`
`impliesa+c=121or-22` For `a+c=-22,` we get `a=c=-11`
Thus, we have `a+c=121`.
`:.a+b+c+d=10(a+c)=1210`