Q187 / 270 · NDA 2025 (Phase 2) · Mathematics
For the following two (02) items:
Let $f(x) = ax^2 + bx + c$ be a quadratic polynomial such that $f(1) = f(4) = 2$. Further, 2 is a root of $f(x) = 0$.
What is the other root of $f(x) = 0$?
Let $f(x) = ax^2 + bx + c$ be a quadratic polynomial such that $f(1) = f(4) = 2$. Further, 2 is a root of $f(x) = 0$.
What is the other root of $f(x) = 0$?
- A$1$
- B$2$
- Correct answerC$3$
- DCannot be determined
Solution
Given $f(1) = f(4) = 2$, the quadratic polynomial can be written as $f(x) = a(x-1)(x-4) + 2$.
Since $2$ is a root of $f(x) = 0$, we have $f(2) = 0$.
Substituting $x = 2$ in the expression for $f(x)$:
$a(2-1)(2-4) + 2 = 0$
$a(1)(-2) + 2 = 0$
$-2a + 2 = 0 \Rightarrow a = 1$
Thus, the polynomial is $f(x) = 1(x-1)(x-4) + 2 = x^2 - 5x + 6$.
To find the roots of $f(x) = 0$, we solve $x^2 - 5x + 6 = 0$.
$(x-2)(x-3) = 0$
The roots are $x = 2$ and $x = 3$.
The other root is $3$.
Answer: $3$
Since $2$ is a root of $f(x) = 0$, we have $f(2) = 0$.
Substituting $x = 2$ in the expression for $f(x)$:
$a(2-1)(2-4) + 2 = 0$
$a(1)(-2) + 2 = 0$
$-2a + 2 = 0 \Rightarrow a = 1$
Thus, the polynomial is $f(x) = 1(x-1)(x-4) + 2 = x^2 - 5x + 6$.
To find the roots of $f(x) = 0$, we solve $x^2 - 5x + 6 = 0$.
$(x-2)(x-3) = 0$
The roots are $x = 2$ and $x = 3$.
The other root is $3$.
Answer: $3$
Hence, the correct answer is option: 'C'.