Question

Assertion (A) : The polynomial $p(y) = y^2 + 4y + 3$ has two zeroes.
Reason (R) : A quadratic polynomial can have at most two zeroes.

• A. Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
• B. Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
• C. Assertion (A) is true, but Reason (R) is false.
• D. Assertion (A) is false, but Reason (R) is true.

Hint:

When solving Assertion-Reason questions, read both statements as independent true/false statements first.
If both are true, insert the word "because" between them:
"The polynomial $y^2 + 4y + 3$ has two zeroes because a quadratic polynomial can have at most two zeroes."
This helping statement sounds incomplete, as "at most two" also includes the possibility of having 1 or 0 zeroes.
Since the general limit doesn't guarantee the exact count of 2 for this specific equation, it is not the correct explanation.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The topic is Polynomials, specifically quadratic polynomials and their zeroes.
We need to analyze:

• Assertion (A) regarding the number of zeroes of the specific polynomial $p(y) = y^2 + 4y + 3$.

• Reason (R) regarding the general maximum limit of zeroes for any quadratic polynomial.


Step 2: Key Formula or Approach:
For a quadratic polynomial $ax^2 + bx + c$:

• The maximum number of real zeroes is equal to its degree, which is 2. Thus, it can have at most two zeroes.

• The actual number of real zeroes depends on the discriminant $D = b^2 - 4ac$.

• If $D \gt 0$, it has two distinct real zeroes.

• If $D = 0$, it has two equal real zeroes (often treated as one distinct zero).

• If $D \lt 0$, it has no real zeroes.


Step 3: Detailed Explanation:

• Analysis of Assertion (A):
The given polynomial is $p(y) = y^2 + 4y + 3$.
To find the zeroes, we set $p(y) = 0$:
\[ y^2 + 4y + 3 = 0 \]
Factorize the quadratic expression by splitting the middle term:
\[ y^2 + 3y + y + 3 = 0 \]
\[ y(y + 3) + 1(y + 3) = 0 \]
\[ (y + 1)(y + 3) = 0 \]
This gives the zeroes:
\[ y = -1 \quad \text{and} \quad y = -3 \]
Since we found exactly two distinct real numbers that make the polynomial zero, the polynomial has two zeroes.
Thus, Assertion (A) is

true.

• Analysis of Reason (R):
According to the fundamental theorem of algebra, a polynomial of degree $n$ can have at most $n$ zeroes.
Since a quadratic polynomial is of degree 2, it can have at most 2 zeroes.
Thus, the statement in Reason (R) is

true.

• Evaluating the Connection:
Although both statements are true, the Reason states a general upper limit ("at most two zeroes").
It does not explain why this specific polynomial $p(y) = y^2 + 4y + 3$ has exactly two zeroes rather than one or zero.
The reason $p(y)$ has exactly two distinct zeroes is because its discriminant $D = b^2 - 4ac = 4^2 - 4(1)(3) = 16 - 12 = 4$ is strictly greater than 0 ($D \gt 0$).
Therefore, Reason (R) is not the complete or direct explanation for Assertion (A).


Step 4: Final Answer:
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). This corresponds to option (B).