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Let a and b be the roots of the equation...

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

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The correct Answer is:
1210
We have `a+b=10c,ab=-11d`
and `c+d=10a,cd=-11b`
`:.a+b+c+d=10(a+c)`
and `abcd=121bd`
`impliesb+d=9(a+c)`
and `ac=121`
Next `a^(2)-10ac=11d=0`
and `c^(2)-10ac-11b=0`
`impliesa^(2)+c^(2)-20ac-11(b+d)=0`
`implies (a+c)^(2)-22xx121-99(a+c)=0`
`impliesa+c=121` or `-22`
IF `a+c=-22impliesa=c` rejecting these values ow have
`a+c=121`
`:.a+b+c=d=10(a+c)=1210`
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