To solve the quadratic polynomial 6x² – 7x – 3, we will find its zeros and verify the relationship between the zeros and the coefficients.
Step 1: Given Quadratic Polynomial
The given polynomial is:
f(x) = 6x² – 7x – 3
Step 2: Finding the Zeros
To find the zeros, set f(x) = 0:
6x² – 7x – 3 = 0
Factorizing the equation:
6x² – 9x + 2x – 3 = 0
Group the terms:
3x(2x – 3) + 1(2x – 3) = 0
Take the common factor:
(2x – 3)(3x + 1) = 0
Set each factor to zero:
- 2x – 3 = 0 → x = 3/2
- 3x + 1 = 0 → x = -1/3
Thus, the zeros of the polynomial are x = 3/2 and x = -1/3.
Step 3: Verifying the Relationship Between Zeros and Coefficients
For a quadratic equation of the form ax² + bx + c, the following relationships hold:
- Sum of Zeros = -b/a
- Product of Zeros = c/a
Here:
- Coefficient of x² (a) = 6
- Coefficient of x (b) = -7
- Constant term (c) = -3
Sum of Zeros:
Zeros are 3/2 and -1/3.
Sum = 3/2 + (-1/3)
To simplify, take the LCM of 2 and 3:
(9 – 2)/6 = 7/6
-b/a = -(-7)/6 = 7/6
Thus, the sum of the zeros matches -b/a.
Product of Zeros:
Product = 3/2 × (-1/3) = -3/6 = -1/2
c/a = -3/6 = -1/2
Thus, the product of the zeros matches c/a.
Final Answer:
The zeros of the quadratic polynomial 6x² – 7x – 3 are x = 3/2 and x = -1/3.
The relationships between the zeros and the coefficients are verified as:
- Sum of zeros = -b/a
- Product of zeros = c/a
Both relationships hold true, confirming the calculations.
