Given the polynomial [tex]\( p(x) = kx^2 - 30x + 45k \)[/tex], we need to find the value of [tex]\( k \)[/tex] such that the zeroes [tex]\( \alpha \)[/tex] and [tex]\( \beta \)[/tex] of the polynomial satisfy [tex]\( \alpha + \beta = \alpha \beta \)[/tex].
Step 1: Identify the coefficients
The polynomial is in the form [tex]\( ax^2 + bx + c \)[/tex], where:
- [tex]\( a = k \)[/tex]
- [tex]\( b = -30 \)[/tex]
- [tex]\( c = 45k \)[/tex]
Step 2: Use properties of polynomial zeroes
For a quadratic polynomial [tex]\( ax^2 + bx + c \)[/tex], the sum and product of zeroes are given by:
- Sum of zeroes [tex]\( \alpha + \beta = -\frac{b}{a} \)[/tex]
- Product of zeroes [tex]\( \alpha \beta = \frac{c}{a} \)[/tex]
Step 3: Calculate the sum and product of zeroes
- Using the formula for the sum of zeroes:
[tex]\[
\alpha + \beta = -\frac{b}{a} = -\left(\frac{-30}{k}\right) = \frac{30}{k}
\][/tex]
- Using the formula for the product of zeroes:
[tex]\[
\alpha \beta = \frac{c}{a} = \frac{45k}{k} = 45
\][/tex]
Step 4: Apply the given condition
We know from the problem statement that [tex]\( \alpha + \beta = \alpha \beta \)[/tex]. Therefore:
[tex]\[
\frac{30}{k} = 45
\][/tex]
Step 5: Solve for [tex]\( k \)[/tex]
To find [tex]\( k \)[/tex]:
[tex]\[
\frac{30}{k} = 45 \implies 30 = 45k \implies k = \frac{30}{45} = \frac{2}{3}
\][/tex]
Conclusion
The value of [tex]\( k \)[/tex] is:
[tex]\[
\boxed{\frac{2}{3}}
\][/tex]
