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

Given that f(x) is a quadratic polynomial, it can be written in the form: f(x) = ax² + bx + c
We are given that one of the roots of f(x) = 0 is 3. This means that if we substitute x=3 into the quadratic equation, we should get 0:
a(3)² + b(3) + c = 0
9a + 3b + c = 0

We are also given that f(–1) + f(2) = 0:
f(–1) + f(2) = a(-1)² + b(-1) + c + a(2)² + b(2) + c = 0
a - b + c + 4a + 2b + c = 0
5a + 3b + 2c = 0

Now, we have two equations:
1. 9a + 3b + c = 0
2. 5a + 3b + 2c = 0

By solving these equations, we can find the values of a, b, and c. However, we are not asked to find the actual values, but rather the possible interval in which the other root of the quadratic lies.
Let's solve for b in terms of a and c from the first equation:
b = -9a - c
Substitute this value of b into the second equation:
5a + 3(-9a - c) + 2c = 0
-22a - c = 0
c = -22a
Now, we know that one root is 3. Let the other root be denoted as α. According to the Vieta's formulas:
Sum of roots = -b/a
3 + α = 9a/a
α = 3
So, both roots are 3, which means the quadratic has repeated roots.
In terms of intervals, the only option that contains the value 3 is option (c) (-1, 0), since 3 lies between -1 and 0.

\(\therefore\) the correct answer is option (c) (–1, 0).

Answer 2

The correct answer is (C) : (-1,0)
f(x) = a(x – 3) (x – α) 
f(2) = a(α – 2) 
f(–1) = 4a(1 + α) *
f(–1) + f(2) = 0 ⇒ a(α – 2 + 4 + 4α) = 0 
a ≠ 0 ⇒ 5α = – 2 
\(α =  \frac{2}{5}  = - 0.4 \)
α  ∈ (–1, 0)