Answer:
Let's address each of the questions step by step:
### (i) If \( a \) and \( \frac{1}{a} \) are the zeroes of the quadratic polynomial \( 2x^2 - rx + 8k \), then find the value of \( k \).
Given:
- Zeroes of the polynomial are \( a \) and \( \frac{1}{a} \).
- The polynomial is \( 2x^2 - rx + 8k \).
For a quadratic polynomial \( ax^2 + bx + c \) with zeroes \( \alpha \) and \( \beta \):
- Sum of the zeroes (\(\alpha + \beta\)) = \(-\frac{b}{a}\).
- Product of the zeroes (\(\alpha \beta\)) = \(\frac{c}{a}\).
Here, \(\alpha = a\) and \(\beta = \frac{1}{a}\).
1. **Sum of the zeroes:**
\[ a + \frac{1}{a} = -\frac{-r}{2} = \frac{r}{2} \]
2. **Product of the zeroes:**
\[ a \cdot \frac{1}{a} = 1 = \frac{8k}{2} \]
\[ 1 = 4k \]
\[ k = \frac{1}{4} \]
So, the value of \( k \) is \( \frac{1}{4} \).
### (ii) Find the sum of zeroes of \( p(x) = kx^2 - kx + 5 \).
Given the polynomial \( p(x) = kx^2 - kx + 5 \), the sum of the zeroes can be found using:
\[ \text{Sum of the zeroes} = -\frac{b}{a} \]
Here:
- \( a = k \)
- \( b = -k \)
So:
\[ \text{Sum of the zeroes} = -\frac{-k}{k} = \frac{k}{k} = 1 \]
The sum of the zeroes is \( 1 \).
### (iii) Write a quadratic polynomial whose one zero is 4 and the product of zeroes is 0.
Let the zeroes be \( 4 \) and \( \beta \).
Given:
- One zero is \( 4 \).
- Product of zeroes is \( 0 \).
Since the product of zeroes is \( 0 \):
\[ 4 \cdot \beta = 0 \]
\[ \beta = 0 \]
So the zeroes are \( 4 \) and \( 0 \).
The quadratic polynomial with zeroes \( \alpha \) and \( \beta \) is given by:
\[ p(x) = k(x - \alpha)(x - \beta) \]
Here:
\[ \alpha = 4 \]
\[ \beta = 0 \]
Thus, the polynomial is:
\[ p(x) = k(x - 4)(x - 0) \]
\[ p(x) = kx(x - 4) \]
For simplicity, let \( k = 1 \):
\[ p(x) = x(x - 4) \]
\[ p(x) = x^2 - 4x \]
The quadratic polynomial is \( x^2 - 4x \).
### (iv) Find the zeroes of \( p(x) = x^2 - 7x + 12 \).
To find the zeroes of \( p(x) = x^2 - 7x + 12 \), we solve the quadratic equation:
\[ x^2 - 7x + 12 = 0 \]
We can factorize this quadratic equation:
\[ x^2 - 7x + 12 = (x - 3)(x - 4) = 0 \]
Thus, the zeroes are:
\[ x - 3 = 0 \Rightarrow x = 3 \]
\[ x - 4 = 0 \Rightarrow x = 4 \]
The zeroes of the polynomial \( p(x) = x^2 - 7x + 12 \) are \( 3 \) and \( 4 \).
