Question

The zeroes of the quadratic polynomial \(x^2+24x+119\) are

• A. one positive and one negative
• B. both positive
• C. both negative
• D. none of the above

Answer 1

The Correct Option is C

To solve the problem, we need to determine the nature (sign) of the zeroes of the quadratic polynomial \( x^2 + 24x + 119 \).

1. Recall the standard form of a quadratic equation:
A quadratic polynomial can be written as:

\( ax^2 + bx + c \)
For our equation, \( a = 1 \), \( b = 24 \), and \( c = 119 \).

2. Find the discriminant:
The discriminant \( D \) of a quadratic equation is given by:

\( D = b^2 - 4ac \)
Substitute the values:

\( D = 24^2 - 4(1)(119) = 576 - 476 = 100 \)
Since \( D > 0 \), the roots are real and distinct.

3. Find the roots using factorization:
We need to find two numbers that multiply to 119 and add to 24.
The numbers are 17 and 7, so we can factor the equation as:

\( x^2 + 17x + 7x + 119 = 0 \)
\( x(x + 17) + 7(x + 17) = 0 \)
\( (x + 7)(x + 17) = 0 \)

Therefore, the roots are \( x = -7 \) and \( x = -17 \).

4. Determine the signs of the roots:
Both roots are negative.

Final Answer:
The correct answer is option (C): both negative.