Certainly! Let's solve each question step-by-step.
### Question 6:
Given:
The quadratic polynomial is [tex]\( p(x) = 2x^2 - 7x + k \)[/tex].
It is given that the zeroes (roots) of the polynomial are reciprocal of each other.
Solution:
1. Let's denote the roots by [tex]\( \alpha \)[/tex] and [tex]\( \frac{1}{\alpha} \)[/tex].
2. For a quadratic polynomial [tex]\( ax^2 + bx + c \)[/tex], the product of the roots [tex]\(\alpha \cdot \frac{1}{\alpha}\)[/tex] is given by [tex]\(\frac{c}{a}\)[/tex].
Here, [tex]\( a = 2 \)[/tex] and [tex]\( c = k \)[/tex].
3. Therefore, the product of the roots is:
[tex]\[
\alpha \cdot \frac{1}{\alpha} = 1 = \frac{k}{2}
\][/tex]
4. Solving for [tex]\( k \)[/tex]:
[tex]\[
1 = \frac{k}{2} \implies k = 2
\][/tex]
So, the value of [tex]\( k \)[/tex] is [tex]\( 2 \)[/tex].
### Question 7:
Given:
The quadratic polynomial is [tex]\( x^2 + (a+1)x + b \)[/tex].
The zeroes (roots) of the polynomial are [tex]\( 2 \)[/tex] and [tex]\( -3 \)[/tex].
Solution:
1. For a quadratic polynomial [tex]\( x^2 + bx + c \)[/tex]:
- The sum of the roots is given by [tex]\( -\frac{b}{a} \)[/tex].
- The product of the roots is given by [tex]\( \frac{c}{a} \)[/tex].
Given roots are [tex]\( 2 \)[/tex] and [tex]\( -3 \)[/tex].
2. Sum of the roots:
[tex]\[
2 + (-3) = -1
\][/tex]
The coefficient of [tex]\( x \)[/tex] here is [tex]\( a + 1 \)[/tex]. So, we have:
[tex]\[
a + 1 = -(-1) \implies a + 1 = 1 \implies a = 0
\][/tex]
3. Product of the roots:
[tex]\[
2 \cdot (-3) = -6
\][/tex]
The constant term, [tex]\( b \)[/tex], is the product of the roots:
[tex]\[
b = -6
\][/tex]
So, the values are [tex]\( a = 0 \)[/tex] and [tex]\( b = -6 \)[/tex].
### Summary:
1. The value of [tex]\( k \)[/tex] is [tex]\( 2 \)[/tex].
2. The values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are [tex]\( 0 \)[/tex] and [tex]\( -6 \)[/tex], respectively.
