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Question

If x210ax11b=0 has the roots c and d, then x210x11d=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 1210
Given: c,d are roots of x210ax11b=0
c+d=10a ...... (1)
cd=11b ...... (2)
Now (1)×c and using equation (2) we get,
c210ac11b=0.......(3)

Also, a,b are roots of x210cx11d=0
a+b=10c.... (4)
ab=11d .... (5)
Multiplying the equation(4) by a and using the 5th equation we get,
a210ac11d=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=121

Now, eq(3)eq.(6)(c2a2)11(bd)=0
c2a2=11(bd)

eq.(1)+eq.(2)a2+c220ac11(b+d)=0
Using the identity (a+b)2=a2+b2+2ab and eq(7) in the above eq. we get,
(a+c)222ac99(a+c)=0
(a+c)299(a+c)2662=0 [ac=121]
(a+c)2(12122)(a+c)121×22=0
(a+c)2121(a+c)+22(a+c)121×22=0
(a+c)(a+c121)+22(a+c121)=0
(a+c+22)(a+c121)=0
a+c=121
a+b+c+d=10(a+c)
a+b+c+d=1210

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