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Theory of Equations: Quadratic Equations Common Root Value 5ab

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

Question
If the quadratic equations 3x2+ax+1=03x^2 + ax + 1 = 0 and 2x2+bx+1=02x^2 + bx + 1 = 0 have a common root, then the value of ∣2a2−5ab+3b2∣|2a^2 - 5ab + 3b^2| is, where 2a≠3b2a \neq 3b
A00
B11correct
C22
D33
Solution
Step 1: Let rr be the common root. Then:
3r2+ar+1=0⋯(1),2r2+br+1=0⋯(2)3r^2 + ar + 1 = 0 \quad \cdots (1), \qquad 2r^2 + br + 1 = 0 \quad \cdots (2)
 Step 2: Multiply (1) by 2 and (2) by 3, then subtract:
(2a−3b)r+(2−3)=0  ⟹  r=13b−2a(2a - 3b)r + (2-3) = 0 \implies r = \frac{1}{3b-2a}
 Step 3: Substituting back into (2) and simplifying using the cross-ratio condition gives:
(3b−2a)(a−b)=1(3b - 2a)(a - b) = 1
 Step 4: Observe that 2a2−5ab+3b2=(2a−3b)(a−b)=−(3b−2a)(a−b)=−12a^2 - 5ab + 3b^2 = (2a-3b)(a-b) = -(3b-2a)(a-b) = -1.
∣2a2−5ab+3b2∣=1|2a^2 - 5ab + 3b^2| = 1
 Correct answer: (2)
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