Question

If the sum of the squares of the roots of the equation $x^2 - (a-2)x - (a+1) = 0$ is least for an appropriate value of the variable parameter $a$, then that value of $a$ will be

• A. 3
• B. 2
• C. 1
• D. 0

Hint:

When asked to minimize expressions involving roots, express everything in terms of sum and product of roots and simplify. For quadratics, minimum or maximum occurs at $a = -\frac{b}{2a}$.

Answer 1

The Correct Option is C

We are given the quadratic equation:

\(x^2 - (a - 2)x - (a + 1) = 0\)

Step 1: General Form of the Quadratic Equation

The given quadratic equation is in the form:

\(x^2 + px + q = 0\)

where $p = -(a - 2) = -a + 2$ and $q = -(a + 1) = -a - 1$.

Step 2: Use of Vieta's Formulas

For the equation $x^2 + px + q = 0$, Vieta’s formulas give the relationships between the coefficients and the roots. Let the roots be $\alpha$ and $\beta$. According to Vieta’s formulas:

$\alpha + \beta = -p$ and $\alpha \beta = q$

In our case:

$\alpha + \beta = a - 2$ and $\alpha \beta = -a - 1$

Step 3: Sum of the Squares of the Roots

We are asked to find the sum of the squares of the roots, which is given by:

\(\alpha^2 + \beta^2\)

Using the identity:

\(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta\)

Substitute the values from Vieta's formulas:

\(\alpha^2 + \beta^2 = (a - 2)^2 - 2(-a - 1)\)

Simplifying:

\(\alpha^2 + \beta^2 = (a - 2)^2 + 2(a + 1)\)

Expanding:

\(\alpha^2 + \beta^2 = (a^2 - 4a + 4) + 2a + 2 = a^2 - 2a + 6\)

Step 4: Minimize the Expression

Now, we need to minimize the expression $a^2 - 2a + 6$. This is a quadratic expression, and we know that a quadratic function of the form $ax^2 + bx + c$ is minimized at $x = -\frac{b}{2a}$.

For our expression $a^2 - 2a + 6$, we have $a = 1$, $b = -2$, and $c = 6$. The value of $a$ that minimizes the expression is:

\(a = -\frac{-2}{2(1)} = 1\)

Step 5: Conclusion

The value of $a$ that minimizes the sum of the squares of the roots is $a = 1$.