Question

A quadratic polynomial whose zeroes are -3 and 4 is :

• A. \(x^2 - x + 12\)
• B. \(x^2 + x + 12\)
• C. \(\frac{x^2}{2} - \frac{x}{2} - 6\)
• D. \(2x^2 + 2x - 24\)

Hint:

If zeroes are \(\alpha\) and \(\beta\): 1. Sum of zeroes: \(S = \alpha + \beta\). 2. Product of zeroes: \(P = \alpha\beta\). 3. The quadratic polynomial is \(k(x^2 - Sx + P)\), where \(k \neq 0\). Given zeroes -3 and 4: \(S = -3 + 4 = 1\) \(P = (-3)(4) = -12\) So, a basic polynomial is \(x^2 - (1)x + (-12) = x^2 - x - 12\). Option (3) is \(\frac{x^2}{2} - \frac{x}{2} - 6 = \frac{1}{2}(x^2 - x - 12)\). This is a constant multiple (\(k=1/2\)) of our basic polynomial, so it has the same zeroes.

Answer 1

The Correct Option is C

Concept: If \(\alpha\) and \(\beta\) are the zeroes of a quadratic polynomial, then the polynomial can be written in the form \(k(x^2 - (\alpha + \beta)x + \alpha\beta)\), where \(k\) is any non-zero constant. Step 1: Identify the given zeroes Let the zeroes be \(\alpha = -3\) and \(\beta = 4\). Step 2: Calculate the sum of the zeroes (\(\alpha + \beta\)) \[ \alpha + \beta = -3 + 4 = 1 \] Step 3: Calculate the product of the zeroes (\(\alpha\beta\)) \[ \alpha\beta = (-3) \times (4) = -12 \] Step 4: Form a quadratic polynomial using the sum and product of zeroes Using the form \(x^2 - (\text{sum of zeroes})x + (\text{product of zeroes})\), one such polynomial is: \[ P(x) = x^2 - (1)x + (-12) \] \[ P(x) = x^2 - x - 12 \] Any non-zero constant multiple of this polynomial will also have the same zeroes. So, any polynomial of the form \(k(x^2 - x - 12)\) where \(k \neq 0\) is a valid answer. Step 5: Check the given options to see which one matches this form or is a multiple
(1) \(x^2 - x + 12\): The constant term is +12, but we need -12. Incorrect.
(2) \(x^2 + x + 12\): Both the coefficient of x and the constant term are incorrect. Incorrect.
(3) \(\frac{x^2}{2} - \frac{x}{2} - 6\): This can be rewritten by factoring out \(\frac{1}{2}\): \[ \frac{1}{2} (x^2 - x - 12) \] This is in the form \(k(x^2 - x - 12)\) with \(k = \frac{1}{2}\). Therefore, this polynomial has zeroes -3 and 4. This option is correct.
(4) \(2x^2 + 2x - 24\): This can be rewritten by factoring out 2: \[ 2(x^2 + x - 12) \] The coefficient of x is +1, but we need -1 inside the parenthesis. Incorrect. (If it were \(2x^2 - 2x - 24\), it would be \(2(x^2 - x - 12)\) and also correct). Thus, option (3) is a quadratic polynomial whose zeroes are -3 and 4.