Question

Let f(x) be a quadratic polynomial such that f(–1) + f(2) = 0. If one of the roots of f(x) = 0 is 3, then its other root lies in :

• A. (-3,-1)
• B. (1,3)
• C. (-1,0)
• D. (0,1)

Answer 1

The Correct Option is C

To solve this problem, let's analyze the given conditions and find out the other root of the quadratic polynomial \( f(x) \).

We know that if \( f(x) \) is a quadratic polynomial, it can be expressed in the form:

f(x) = a(x - \alpha)(x - \beta)

where \( \alpha \) and \( \beta \) are the roots of the polynomial.

Given that one of the roots is \( 3 \), we have:

\alpha = 3

Thus, the polynomial can be expressed as:

f(x) = a(x - 3)(x - \beta)

It is also given that \( f(-1) + f(2) = 0 \). We need to find \( \beta \) such that satisfies this condition.

Step 1: Calculate \( f(-1) \)

f(-1) = a(-1 - 3)(-1 - \beta) = a(-4)(-1 - \beta) = 4a(1 + \beta)

Step 2: Calculate \( f(2) \)

f(2) = a(2 - 3)(2 - \beta) = a(-1)(2 - \beta) = -a(2 - \beta)

Step 3: Use the condition \( f(-1) + f(2) = 0 \)

4a(1 + \beta) - a(2 - \beta) = 0

Solving the equation:

\Rightarrow 4a + 4a\beta - 2a + a\beta = 0 \Rightarrow a(4\beta + \beta + 4 - 2) = 0 \Rightarrow a(5\beta + 2) = 0

Since \( a \neq 0 \) (otherwise the polynomial will not have a quadratic form),

5\beta + 2 = 0 \Rightarrow \beta = -\frac{2}{5}

Conclusion

The other root \( \beta \) is \( -\frac{2}{5} \), which lies in the interval \((-1, 0)\).

Therefore, the correct answer is: (-1, 0)