Question:medium

The value of 'a' so that the sum of squares of the roots of the equation $x^2 - (a - 2)x - a + 1 = 0$ assumes the least value is

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To minimize a quadratic expression $Ax^2+Bx+C$, find the vertex at $x = -B/(2A)$.
Updated On: May 14, 2026
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
We are given a quadratic equation parameterized by '$a$'. We need to express the sum of the squares of its roots in terms of '$a$' and then find the value of '$a$' that minimizes this expression.
Step 2: Key Formula or Approach:
For a quadratic equation $Ax^2 + Bx + C = 0$ with roots $\alpha$ and $\beta$:
Sum of roots: $\alpha + \beta = -\frac{B}{A}$
Product of roots: $\alpha \beta = \frac{C}{A}$
Sum of squares of roots: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$.
To minimize a quadratic expression $f(a)$, complete the square or find the vertex.
Step 3: Detailed Explanation:
The given equation is $x^2 - (a - 2)x - (a - 1) = 0$.
Let the roots be $\alpha$ and $\beta$.
Sum of the roots ($\alpha + \beta$) is the negative coefficient of $x$:
\[ \alpha + \beta = -(-(a - 2)) = a - 2 \] Product of the roots ($\alpha \beta$) is the constant term:
\[ \alpha \beta = -(a - 1) = -a + 1 = 1 - a \] We need to minimize the sum of squares, let's call it $S$:
\[ S = \alpha^2 + \beta^2 \] Using algebraic identity:
\[ S = (\alpha + \beta)^2 - 2\alpha \beta \] Substitute the sum and product:
\[ S = (a - 2)^2 - 2(1 - a) \] Expand the terms:
\[ S = (a^2 - 4a + 4) - 2 + 2a \] Combine like terms to form a quadratic in terms of '$a$':
\[ S = a^2 - 2a + 2 \] To find the minimum value, we can complete the square:
\[ S = (a^2 - 2a + 1) + 1 \] \[ S = (a - 1)^2 + 1 \] Since the square of any real number is non-negative ($(a - 1)^2 \ge 0$), the minimum value of $S$ occurs when the squared term is zero.
\[ (a - 1)^2 = 0 \implies a = 1 \] At $a=1$, the minimum value of the sum of squares is $1$.
Step 4: Final Answer:
The value of 'a' for the least sum of squares is 1.
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