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✅ Step-by-Step Solution
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Polynomials Class 10 🔥 If α, β are zeroes of x² – 5x + k & α – β = 1, find k
📝 Key Formulas Used
Key Concepts:
• Sum of zeroes: α + β = –b/a
• Product of zeroes: αβ = c/a
• Standard quadratic form: ax² + bx + c = 0
• Solving pair of linear equations
⚠️ Common Mistakes Students Make
- Forgetting to add the two equations — students try to solve α and β separately. Always add (α+β) and (α–β) together to eliminate one variable first.
- Wrong sign in sum formula — α+β = –b/a, not b/a. In x² – 5x + k, b = –5, so –b/a = –(–5)/1 = 5.
- Not verifying the answer — always put α and β back in the polynomial to confirm both give zero.
💡 Alternate Method
From α + β = 5 and α – β = 1, directly find: α = ((α+β) + (α–β)) / 2 = (5 + 1) / 2 = 3, β = ((α+β) – (α–β)) / 2 = (5 – 1) / 2 = 2. Then k = αβ = 3 × 2 = 6.
Verification: For α = 3: (3)² – 5(3) + 6 = 9 – 15 + 6 = 0 ✅. For β = 2: (2)² – 5(2) + 6 = 4 – 10 + 6 = 0 ✅
Teacher's Note
This solution has been carefully prepared, but we recommend showing it to your Maths teacher once — marking schemes can vary slightly between boards (CBSE, ICSE, State Boards). Your teacher knows exactly what your examiner expects.
📝 Practice Questions (Try Yourself)
Q1. If α and β are zeroes of x² – 6x + k such that α – β = 2, find k.
Show Answer
k = 8 (α = 4, β = 2)
Q2. If α and β are zeroes of 2x² + 5x + k such that α + β + αβ = –1, find k.
Show Answer
k = –3 (α+β = –5/2, αβ = k/2 → –5/2 + k/2 = –1 → k = –3)
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