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Theory of Equations: Distinct Positive Real Numbers Roots 10cx 11d Roots

JEE Maths question with a full step-by-step solution.

Question
If a,b,c,da, b, c, d are distinct positive real numbers such that aa and bb are the roots of x2−10cx−11d=0x^2 - 10cx - 11d = 0 and cc and dd are the roots of x2−10ax−11b=0x^2 - 10ax - 11b = 0, then the value of a+b+c+da + b + c + d is
A11101110
B10101010
C11011101
D12101210correct
Solution
Step 1: By Vieta's formulas:
a+b=10c,ab=−11d⋯(1),c+d=10a,cd=−11b⋯(2)a+b = 10c, \quad ab = -11d \quad \cdots (1), \qquad c+d = 10a, \quad cd = -11b \quad \cdots (2)
 Step 2: Since aa is a root of the first equation and cc is a root of the second
a2−10ca−11d=0,c2−10ac−11b=0a^2 - 10ca - 11d = 0, \quad c^2 - 10ac - 11b = 0
 Subtracting: (a2−c2)−11(d−b)=0(a^2 - c^2) - 11(d - b) = 0, and from (1): d=10a−cd = 10a - c, b=10c−ab = 10c - a. Step 3: Substituting and simplifying gives:
(a−c)[−(a+c)+121]=0(a-c)\bigl[-(a+c) + 121\bigr] = 0
 Since a≠ca \neq c (distinct), a+c=121a + c = 121. Step 4:
a+b+c+d=(a+b)+(c+d)=10c+10a=10(a+c)=10×121=1210a + b + c + d = (a + b) + (c + d) = 10c + 10a = 10(a + c) = 10 \times 121 = 1210
 Correct answer: (4)
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