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Theory of Equations: Non Zero Real Numbers Roots Roots Absolute Value

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

Question
If a,b,c,da, b, c, d are non-zero real numbers such that cc and dd are the roots of x2+ax+b=0x^2 + ax + b = 0 and aa and bb are the roots of x2+cx+d=0x^2 + cx + d = 0, then the absolute value of a+2b+3c+4da + 2b + 3c + 4d is
A22
B44
C66
D88correct
Solution
Step 1: From the two equations:
c+d=−a,cd=b⋯(1),a+b=−c,ab=d⋯(2)c + d = -a, \quad cd = b \quad \cdots (1), \qquad a + b = -c, \quad ab = d \quad \cdots (2)
 Step 2: From c+d=−ac+d=-a and a+b=−ca+b=-c: adding gives a+b+c+d=−(a+c)a+b+c+d = -(a+c), so b+d=−2(a+c)b+d = -2(a+c). From c+d=−ac+d=-a: d−b=0−b−c−...d-b=0-b-c-... Let me use direct algebra. From (1) and (2): c+d=−ac+d=-a ... (I) and a+b=−ca+b=-c ... (II). Subtracting: c+d−a−b=−a+c  ⟹  d−b=0  ⟹  d=bc+d-a-b = -a+c \implies d-b=0 \implies d = b. Step 3: d=b  ⟹  ab=d=b  ⟹  b(a−1)=0d = b \implies ab = d = b \implies b(a-1) = 0. Since b≠0b \neq 0: a=1a = 1. Step 4: From a+b=−ca + b = -c: 1+b=−c1+b = -c, so c=−1−bc = -1-b. From cd=bcd = b: (−1−b)b=b  ⟹  −1−b=1  ⟹  b=−2(-1-b)b = b \implies -1-b = 1 \implies b = -2 (since b≠0b \neq 0). So c=1c = 1, d=−2d = -2. Step 5: ∣a+2b+3c+4d∣=∣1+2(−2)+3(1)+4(−2)∣=∣1−4+3−8∣=∣−8∣=8|a + 2b + 3c + 4d| = |1 + 2(-2) + 3(1) + 4(-2)| = |1-4+3-8| = |-8| = 8. Correct answer: (4)
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