Q1(iii):

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

Given: A quadratic polynomial $p(x) = 6x^2 - 3 - 7x$.

To Find: The zeroes of the polynomial and verify the relationship between the zeroes and the coefficients.

Step 1: Standardizing the Polynomial
The standard form of a quadratic polynomial is $ax^2 + bx + c$. We rearrange the given polynomial in descending order of the powers of $x$:
$p(x) = 6x^2 - 7x - 3$

Step 2: Finding the Zeroes by Factorization
To find the zeroes, we set $p(x) = 0$.
$6x^2 - 7x - 3 = 0$
We use the splitting the middle term method. We look for two numbers whose product is $a \times c = 6 \times (-3) = -18$ and whose sum is $b = -7$.
The factors of $-18$ that sum to $-7$ are $-9$ and $2$.
$6x^2 - 9x + 2x - 3 = 0$
Group the terms:
$(6x^2 - 9x) + (2x - 3) = 0$
Factor out the common terms:
$3x(2x - 3) + 1(2x - 3) = 0$
$(3x + 1)(2x - 3) = 0$

Step 3: Solving for $x$
Setting each factor to zero:
1) $3x + 1 = 0 \implies 3x = -1 \implies x = -\frac{1}{3}$
2) $2x - 3 = 0 \implies 2x = 3 \implies x = \frac{3}{2}$
Thus, the zeroes are $\alpha = \frac{3}{2}$ and $\beta = -\frac{1}{3}$.

Step 4: Verifying the Relationship between Zeroes and Coefficients
For the polynomial $6x^2 - 7x - 3$, the coefficients are $a = 6$, $b = -7$, and $c = -3$.

Verification of Sum of Zeroes:
Formula: $\alpha + \beta = -\frac{b}{a}$
LHS: $\alpha + \beta = \frac{3}{2} + (-\frac{1}{3}) = \frac{9 - 2}{6} = \frac{7}{6}$
RHS: $-\frac{b}{a} = -\frac{-7}{6} = \frac{7}{6}$
Since LHS = RHS, the relationship is verified.

Verification of Product of Zeroes:
Formula: $\alpha \cdot \beta = \frac{c}{a}$
LHS: $\alpha \cdot \beta = (\frac{3}{2}) \cdot (-\frac{1}{3}) = -\frac{3}{6} = -\frac{1}{2}$
RHS: $\frac{c}{a} = \frac{-3}{6} = -\frac{1}{2}$
Since LHS = RHS, the relationship is verified.

Final Answer: The zeroes of the polynomial are $\frac{3}{2}$ and $-\frac{1}{3}$. The relationship between the zeroes and coefficients is verified as the sum of zeroes is $\frac{7}{6}$ and the product of zeroes is $-\frac{1}{2}$.