Question:
The value of \(\lambda\) such that sum of the squares of the roots of the quadratic equation, \(x^2 + (3 - \lambda)x + 2 = \lambda\) has the least value is -
The value of \(\lambda\) such that sum of the squares of the roots of the quadratic equation, \(x^2 + (3 - \lambda)x + 2 = \lambda\) has the least value is -
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Write the equation as \(x^2 + (3-\lambda)x + (2-\lambda) = 0\), use \(\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta\) and minimise the resulting quadratic in \(\lambda\).
Updated On: Nov 5, 2026
- 1
- 2
- \(\frac{15}{8}\)
- \(\frac{4}{9}\)
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The Correct Option is B
Solution and Explanation
The equation is \(x^2 + (3-\lambda)x + (2-\lambda) = 0\).
Sum of roots \(\alpha+\beta = \lambda - 3\); product \(\alpha\beta = 2 - \lambda\).
\(\alpha^2 + \beta^2 = (\lambda-3)^2 - 2(2-\lambda) = \lambda^2 - 6\lambda + 9 - 4 + 2\lambda = \lambda^2 - 4\lambda + 5\).
\(= (\lambda-2)^2 + 1\), which is least (equal to 1) when \(\lambda = 2\).
Answer: (2) 2
Sum of roots \(\alpha+\beta = \lambda - 3\); product \(\alpha\beta = 2 - \lambda\).
\(\alpha^2 + \beta^2 = (\lambda-3)^2 - 2(2-\lambda) = \lambda^2 - 6\lambda + 9 - 4 + 2\lambda = \lambda^2 - 4\lambda + 5\).
\(= (\lambda-2)^2 + 1\), which is least (equal to 1) when \(\lambda = 2\).
Answer: (2) 2
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