Question

Find the zeroes of the quadratic polynomial \( x^2 + 12 + 7x \) and verify the relationship between the zeroes and the coefficients.

Hint:

For quadratic polynomials, use \( \text{Sum of zeroes} = -\frac{b}{a} \) and \( \text{Product of zeroes} = \frac{c}{a} \) to verify relationships quickly.

Answer 1

Step 1: Write the polynomial in standard form.
The given polynomial is: \[ p(x) = x^2 + 7x + 12 \] Here, \( a = 1, b = 7, c = 12 \).
Step 2: Find the zeroes of the polynomial.
We can find the zeroes by factorizing the quadratic expression: \[ x^2 + 7x + 12 = 0 \] We need two numbers whose product is 12 and sum is 7. \[ 3 \times 4 = 12, \quad 3 + 4 = 7 \] Hence, the factors are (x + 3)(x + 4). \[ x^2 + 7x + 12 = (x + 3)(x + 4) \] So, \[ x = -3, \, -4 \]
Step 3: Verify the relationship between zeroes and coefficients.
For a quadratic polynomial \( ax^2 + bx + c \): \[ \text{Sum of zeroes} = -\frac{b}{a}, \quad \text{Product of zeroes} = \frac{c}{a} \] Sum of zeroes: \[ (-3) + (-4) = -7 \] \[ -\frac{b}{a} = -\frac{7}{1} = -7 \quad \text{✓ Verified.} \] Product of zeroes: \[ (-3)(-4) = 12 \] \[ \frac{c}{a} = \frac{12}{1} = 12 \quad \text{✓ Verified.} \]
Step 4: Final Answer.
\[ \boxed{\text{Zeroes are } -3 \text{ and } -4. \text{ Relation verified.}} \]