Multiple choice

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:

  1. $2$
  2. $\dfrac{4}{9}$
  3. $\dfrac{15}{8}$
  4. $1$
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A Correct answer
Explanation

Let the roots be x1, x2. x1+x2 = -(3-lambda) = lambda-3, x1*x2 = 2-lambda. Sum of squares = (x1+x2)^2 - 2(x1*x2) = (lambda-3)^2 - 2(2-lambda) = lambda^2 - 6lambda + 9 - 4 + 2lambda = lambda^2 - 4lambda + 5. This is a parabola opening upward with minimum at lambda = -(-4)/(2*1) = 2.

AI explanation

Using the relationship between roots and coefficients, let the roots be p and q, giving the sum p + q = -(3 - L) and the product pq = 2 - L. The sum of their squares is p^2 + q^2 = (p + q)^2 - 2pq = (L - 3)^2 - 2(2 - L). Expanding and simplifying yields the quadratic expression L^2 - 4L + 5, which represents an upward opening parabola. This quadratic attains its minimum value at the vertex, where L = -(-4) / (2 * 1) = 2.

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