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Question

# If x2−10ax−11b=0 has the roots c and d, then x2−10x−11d=0 has the roots a and b. Find the value of a+b+c+d if a+c is positive.

A
1220
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B
1110
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C
1210
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D
1310
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Solution

## The correct option is C 1210Given: c,d are roots of x2−10ax−11b=0∴c+d=10a ...... (1)cd=−11b ...... (2) Now (1)×c and using equation (2) we get, c2−10ac−11b=0.......(3)Also, a,b are roots of x2−10cx−11d=0a+b=10c.... (4)ab=−11d .... (5)Multiplying the equation(4) by a and using the 5th equation we get,a2−10ac−11d=0.......(6)Now from the above equations we have, a+b+c+d=10(a+c)⇒b+d=9(a+c) ..... (7)abcd=121bd∴ac=121Now, eq(3)−eq.(6)⇒(c2−a2)−11(b−d)=0c2−a2=11(b−d)eq.(1)+eq.(2)⇒a2+c2−20ac−11(b+d)=0Using the identity (a+b)2=a2+b2+2ab and eq(7) in the above eq. we get, (a+c)2−22ac−99(a+c)=0(a+c)2−99(a+c)−2662=0 [∵ac=121](a+c)2−(121−22)(a+c)−121×22=0(a+c)2−121(a+c)+22(a+c)−121×22=0(a+c)(a+c−121)+22(a+c−121)=0(a+c+22)(a+c−121)=0∴a+c=121a+b+c+d=10(a+c)∴a+b+c+d=1210

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