Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
`6x^(2)-3-7x`

Answer

To find the zeroes of the quadratic polynomial \(6x^2 - 7x - 3\) and verify the relationship between the zeroes and the coefficients, we will follow these steps:

### Step 1: Write the polynomial in standard form
The given polynomial is:
\[ 6x^2 - 7x - 3 \]
This is already in standard form \(ax^2 + bx + c\) where:
- \(a = 6\)
- \(b = -7\)
- \(c = -3\)

### Step 2: Identify the coefficients
From the polynomial, we have:
- \(a = 6\)
- \(b = -7\)
- \(c = -3\)

### Step 3: Use the quadratic formula to find the zeroes
The zeroes of a quadratic polynomial can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values of \(a\), \(b\), and \(c\):
\[ x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \cdot 6 \cdot (-3)}}{2 \cdot 6} \]
\[ x = \frac{7 \pm \sqrt{49 + 72}}{12} \]
\[ x = \frac{7 \pm \sqrt{121}}{12} \]
\[ x = \frac{7 \pm 11}{12} \]

### Step 4: Calculate the zeroes
Now we calculate the two possible values for \(x\):
1. \(x_1 = \frac{7 + 11}{12} = \frac{18}{12} = \frac{3}{2}\)
2. \(x_2 = \frac{7 - 11}{12} = \frac{-4}{12} = -\frac{1}{3}\)

Thus, the zeroes of the polynomial are:
\[ x_1 = \frac{3}{2}, \quad x_2 = -\frac{1}{3} \]

### Step 5: Verify the relationship between the zeroes and the coefficients
The sum of the zeroes \(x_1 + x_2\) can be calculated as:
\[ x_1 + x_2 = \frac{3}{2} + \left(-\frac{1}{3}\right) = \frac{3}{2} - \frac{1}{3} \]
To add these fractions, we need a common denominator, which is 6:
\[ x_1 + x_2 = \frac{3 \cdot 3}{6} - \frac{1 \cdot 2}{6} = \frac{9 - 2}{6} = \frac{7}{6} \]

According to the relationship, the sum of the zeroes should equal \(-\frac{b}{a}\):
\[ -\frac{b}{a} = -\frac{-7}{6} = \frac{7}{6} \]

The product of the zeroes \(x_1 \cdot x_2\) can be calculated as:
\[ x_1 \cdot x_2 = \frac{3}{2} \cdot \left(-\frac{1}{3}\right) = -\frac{3}{6} = -\frac{1}{2} \]

According to the relationship, the product of the zeroes should equal \(\frac{c}{a}\):
\[ \frac{c}{a} = \frac{-3}{6} = -\frac{1}{2} \]

### Conclusion
Both the sum and product of the zeroes match the relationships derived from the coefficients:
- Sum of zeroes: \( \frac{7}{6} \) (matches \(-\frac{b}{a}\))
- Product of zeroes: \( -\frac{1}{2} \) (matches \(\frac{c}{a}\))

Thus, we have verified the relationship between the zeroes and the coefficients.