Question

The sum and product of zeroes of a quadratic polynomial $p(x)$ are $-\frac{1}{3}$ and $2$ respectively. The polynomial $p(x)$ is :

• A. $3x^2 - x + 6$
• B. $x^2 + \frac{1}{3}x - 2$
• C. $3x^2 - x + 2$
• D. $-3x^2 - x - 6$

Hint:

For any quadratic polynomial $ax^2 + bx + c$, the sum of zeroes is $-\frac{b}{a}$ and the product is $\frac{c}{a}$.
In a multiple-choice question, you can quickly test this on the options:
For Option (A): Sum $= -\frac{-1}{3} = \frac{1}{3} \neq -\frac{1}{3}$.
For Option (B): Product $= \frac{-2}{1} = -2 \neq 2$.
For Option (C): Sum $= -\frac{-1}{3} = \frac{1}{3} \neq -\frac{1}{3}$.
For Option (D): Sum $= -\frac{-1}{-3} = -\frac{1}{3}$ and Product $= \frac{-6}{-3} = 2$.
This verification process takes only a few seconds and prevents mistakes!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
This question belongs to the topic of "Polynomials".
We are given the sum and product of the zeroes of a quadratic polynomial $p(x)$.
Specifically, let the zeroes be $\alpha$ and $\beta$. We are given:
\[ \alpha + \beta = -\frac{1}{3} \] \[ \alpha\beta = 2 \] We need to find which of the given polynomial equations matches these conditions.

Step 2: Key Formula or Approach:
Any quadratic polynomial $p(x)$ can be written in terms of its sum ($S$) and product ($P$) of zeroes as:
\[ p(x) = k \left( x^2 - Sx + P \right) \] where $k$ is a non-zero real constant.
By substituting $S = -\frac{1}{3}$ and $P = 2$, we obtain a general family of polynomials and then select a value of $k$ that matches one of the given multiple-choice options.

Step 3: Detailed Explanation:

• Define the sum of zeroes ($S$) and product of zeroes ($P$):
\[ S = \alpha + \beta = -\frac{1}{3} \] \[ P = \alpha\beta = 2 \]

• Formulate the polynomial template:
\[ p(x) = k \left( x^2 - Sx + P \right) \]

• Substitute $S$ and $P$ into the template:
\[ p(x) = k \left( x^2 - \left(-\frac{1}{3}\right)x + 2 \right) \] \[ p(x) = k \left( x^2 + \frac{1}{3}x + 2 \right) \]

• Let us evaluate this expression for different values of $k$ to find a match with the options:

• If we choose $k = 3$:
\[ p(x) = 3 \left( x^2 + \frac{1}{3}x + 2 \right) = 3x^2 + x + 6 \] This does not match Option (A) because Option (A) is $3x^2 - x + 6$.

• If we choose $k = -3$:
\[ p(x) = -3 \left( x^2 + \frac{1}{3}x + 2 \right) = -3x^2 - x - 6 \] This perfectly matches Option (D).

• To be absolutely sure, let's verify the sum and product of zeroes for Option (D) $p(x) = -3x^2 - x - 6$:
Here, $a = -3$, $b = -1$, and $c = -6$.
- Sum of zeroes $= -\frac{b}{a} = -\frac{-1}{-3} = -\frac{1}{3}$ (Correct)
- Product of zeroes $= \frac{c}{a} = \frac{-6}{-3} = 2$ (Correct)

Step 4: Final Answer:
The quadratic polynomial is $-3x^2 - x - 6$, which corresponds to Option (D).