Question

Find the quadratic polynomial the sum of whose zeroes is 1 and their product is –12. Hence find the zeroes of the polynomial.

Hint:

Always verify your answers:
Sum of zeroes: \(4 + (-3) = 1\) (Matches given)
Product of zeroes: \(4 \times (-3) = -12\) (Matches given)

Answer 1

Step 1: Understanding the Question:
We are given the sum and product of the zeroes of a quadratic polynomial. We need to construct the polynomial and then find its zeroes.

Step 2: Key Formula or Approach:
- A quadratic polynomial with sum of zeroes \(S\) and product of zeroes \(P\) is given by:
\[ f(x) = k[x^2 - Sx + P] \]
where \(k\) is a non-zero real constant. For simplicity, we can choose \(k = 1\).
- To find the zeroes, solve \(f(x) = 0\) using factorization.

Step 3: Detailed Explanation:

• Given values:
Sum of zeroes, \(S = 1\)
Product of zeroes, \(P = -12\)

• Write down the polynomial:
\[ f(x) = x^2 - (1)x + (-12) = x^2 - x - 12 \]

• To find the zeroes, set \(f(x) = 0\):
\[ x^2 - x - 12 = 0 \]

• Split the middle term:
Find two numbers whose sum is \(-1\) and product is \(-12\). These numbers are \(-4\) and \(3\).
\[ x^2 - 4x + 3x - 12 = 0 \]
\[ x(x - 4) + 3(x - 4) = 0 \]
\[ (x - 4)(x + 3) = 0 \]

• Find the roots:
\[ x - 4 = 0 \implies x = 4 \]
\[ x + 3 = 0 \implies x = -3 \]


Step 4: Final Answer:
The quadratic polynomial is \(x^2 - x - 12\) and its zeroes are 4 and \(-3\).