Multiple choice

If $\alpha,\ \beta$ are roots of the quadratic equation $\mathrm{x}^{2}-\mathrm{x}(\mathrm{a}-2)-(\mathrm{a}+1)=0$ when $\mathrm{a}$ is a variable then the least value of $\alpha^{2}+\beta^{2}$

  1. $5$
  2. $3$
  3. $4$
  4. $1$
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A Correct answer
Explanation

alpha + beta = a-2, alpha*beta = -(a+1). alpha^2 + beta^2 = (alpha+beta)^2 - 2*alpha*beta = (a-2)^2 + 2(a+1) = a^2 - 4a + 4 + 2a + 2 = a^2 - 2a + 6. This is a parabola opening upward with vertex at a = -(-2)/2 = 1. Min value = 1^2 - 2(1) + 6 = 5.

AI explanation

For the quadratic equation x^2 - x(a - 2) - (a + 1) = 0, the sum of the roots alpha + beta equals a - 2 and the product alpha*beta equals -(a + 1). The expression alpha^2 + beta^2 can be rewritten as (alpha + beta)^2 - 2(alpha*beta). Substituting the sum and product gives (a - 2)^2 - 2(-a - 1), which simplifies to a^2 - 4a + 4 + 2a + 2 = a^2 - 2a + 6. Completing the square for this quadratic in a gives (a - 1)^2 + 5, which reaches its minimum value of 5 when a = 1.

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