Common Roots
If the two quadratic equations x^2 + bx + c = 0 and x^2 + cx + b = 0, with b not equal to c, share exactly one common root, then the value of b + c equals which expression?
Select the correct option:
Solution
−1
When two quadratics share a common root, subtracting them eliminates the squared term and isolates that root, the canonical JEE Advanced common-root method. Subtracting the second equation from the first gives (b - c)x + (c - b) = 0, that is (b - c)x - (b - c) = 0. Since b ≠ c we may divide by (b - c) to get x - 1 = 0, so the common root is x = 1. Substituting x = 1 into the first equation gives 1 + b + c = 0, hence b + c = -1. The elimination step turning two quadratics into a linear relation is the key archetype. Option 0 ignores the constant from substitution. Option 1 has the wrong sign. Option cannot be determined overlooks that the common root is forced to be 1. Hence b + c = -1. Plausibility check: with b + c = -1 the common root 1 satisfies both equations identically, and the condition b ≠ c keeps the two quadratics genuinely different, confirming consistency. Eliminating the squared term by subtracting two quadratics is the universal opening move for common-root problems, reducing the system to a single linear equation that pins down the shared root. Substituting that root back into either original equation then recovers the relation among the coefficients, and the condition that the two quadratics remain distinct is what keeps the division step legitimate.
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About This Question
- Subject
- mathematics
- Chapter
- complex numbers and quadratic equations
- Topic
- common roots
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
−1
When two quadratics share a common root, subtracting them eliminates the squared term and isolates that root, the canonical JEE Advanced common-root method. Subtracting the second equation from the first gives (b - c)x + (c - b) = 0, that is (b - c)x - (b - c) = 0. Since b ≠ c we may divide by (b - c) to get x - 1 = 0, so the common root is x = 1. Substituting x = 1 into the first equation gives 1 + b + c = 0, hence b + c = -1. The elimination step turning two quadratics into a linear relation is the key archetype. Option 0 ignores the constant from substitution. Option 1 has the wrong sign. Option cannot be determined overlooks that the common root is forced to be 1. Hence b + c = -1. Plausibility check: with b + c = -1 the common root 1 satisfies both equations identically, and the condition b ≠ c keeps the two quadratics genuinely different, confirming consistency. Eliminating the squared term by subtracting two quadratics is the universal opening move for common-root problems, reducing the system to a single linear equation that pins down the shared root. Substituting that root back into either original equation then recovers the relation among the coefficients, and the condition that the two quadratics remain distinct is what keeps the division step legitimate.
This medium difficulty mathematics question is from the chapter complex numbers and quadratic equations, covering the topic of common roots. It appeared in the 2025 exam.
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