(a) If [tex]\alpha[/tex] and [tex]\beta[/tex] are roots of [tex]f(x)[/tex], then we can factorize [tex]f[/tex] as
[tex]f(x) = x^2 + (k - 6) x + 9 = (x - \alpha) (x - \beta)[/tex]
Expand the right side and match up coefficients:
[tex]x^2 + (k-6) x + 9 = x^2 - (\alpha + \beta) x + \alpha \beta \implies \begin{cases} \alpha + \beta = -(k-6) \\ \alpha \beta = 9 \end{cases}[/tex]
Now, recall that [tex](x+y)^2 = x^2 + 2xy + y^2[/tex]. It follows that
[tex]\boxed{\alpha^2 + \beta^2} = (\alpha + \beta)^2 - 2\alpha\beta = (-(k-6))^2 - 2\times9 = \boxed{k^2 - 12k + 18}[/tex]
and
[tex]\boxed{\alpha^2\beta^2} = 9^2 = \boxed{81}[/tex]
(b) If [tex]9(\alpha^2+\beta^2) = 2\alpha^2\beta^2[/tex], then
[tex]9 (k^2 - 12k + 18) = 2\times81 \implies 9k^2 - 108k = 0 \implies 9k (k - 12) = 0[/tex]
Since [tex]k\neq0[/tex], it follows that [tex]\boxed{k=12}[/tex].
(c) The simplest quadratic expression with roots [tex]\frac1{\alpha^2}[/tex] and [tex]\frac1{\beta^2}[/tex] is
[tex]\left(x - \dfrac1{\alpha^2}\right) \left(x - \dfrac1{\beta^2}\right)[/tex]
which expands to
[tex]x^2 - \left(\dfrac1{\alpha^2} + \dfrac1{\beta^2}\right) x + \dfrac1{\alpha^2\beta^2}[/tex]
Reusing the identity from (a-i) and the result from part (b), we have
[tex]\left(\dfrac1\alpha + \dfrac1\beta\right)^2 = \dfrac1{\alpha^2} + \dfrac2{\alpha\beta} + \dfrac1{\beta^2} \\\\ \implies \dfrac1{\alpha^2} + \dfrac1{\beta^2} = \left(\dfrac{\alpha + \beta}{\alpha\beta}\right)^2 - \dfrac2{\alpha\beta} = \left(\dfrac{-(k-6)}9\right)^2 - \dfrac29 = \dfrac29[/tex]
We also know from part (a-ii) that [tex]\alpha^2\beta^2=81[/tex].
So, the simplest quadratic that fits the description is
[tex]x^2 - \dfrac29 x + \dfrac1{81}[/tex]
To get one with integer coefficients, we multiply the whole expression by 81 to get [tex]\boxed{81x^2 - 18x + 1}[/tex].
