Problem and Solution
The quadratic equation is $$ 3x^2 - kx - 1 = 0 $$ with roots $$ \alpha \quad \text{and} \quad \beta $$
Part (a): Showing $ \alpha^2 + \beta^2 = \frac{k^2 + 6}{9} $
Using the identity
$$ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta $$
and substituting
$$ \alpha + \beta = \frac{k}{3}, \quad \alpha\beta = \frac{-1}{3} $$
we get
$$ \alpha^2 + \beta^2 = \left(\frac{k}{3}\right)^2 - 2\left(\frac{-1}{3}\right) $$
Simplify:
$$ \alpha^2 + \beta^2 = \frac{k^2}{9} + \frac{2}{3} $$
Converting $$ \frac{2}{3} $$ to denominator $9$:
$$ \alpha^2 + \beta^2 = \frac{k^2}{9} + \frac{6}{9} = \frac{k^2 + 6}{9} $$
Part (b): Finding $k$ given $ \alpha^4 + \beta^4 = \frac{466}{81} $
Using the identity
$$ \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2 $$
and substituting
$$ \alpha^2 + \beta^2 = \frac{k^2 + 6}{9}, \quad (\alpha\beta)^2 = \left(\frac{-1}{3}\right)^2 = \frac{1}{9} $$
we have
$$ \alpha^4 + \beta^4 = \left(\frac{k^2 + 6}{9}\right)^2 - 2\left(\frac{1}{9}\right) $$
Simplify:
$$ \alpha^4 + \beta^4 = \frac{(k^2 + 6)^2}{81} - \frac{2}{9} $$
Convert $$ \frac{2}{9} = \frac{18}{81} $$
Thus
$$ \alpha^4 + \beta^4 = \frac{(k^2 + 6)^2}{81} - \frac{18}{81} $$
Equating to $$ \frac{466}{81} $$ we get
$$ \frac{(k^2 + 6)^2}{81} - \frac{18}{81} = \frac{466}{81} $$
Multiply through by $81$:
$$ (k^2 + 6)^2 - 18 = 466 $$
Simplify:
$$ (k^2 + 6)^2 = 484 $$
Take square roots:
$$ k^2 + 6 = \pm 22 $$
Since $k$ is positive:
$$ k^2 + 6 = 22 \implies k^2 = 16 \implies k = 4 $$
Part (c): Forming the quadratic equation
The roots are
$$ r_1 = \frac{\alpha^3 + \beta}{\beta}, \quad r_2 = \frac{\alpha + \beta^3}{\alpha} $$
Step 1: Sum of the roots $ (r_1 + r_2) $
Using
$$ r_1 + r_2 = \frac{\alpha^4 + \beta^4 + 2\alpha\beta}{\alpha\beta} $$
and substituting
$$ \alpha^4 + \beta^4 = \frac{466}{81}, \quad \alpha\beta = \frac{-1}{3} $$
we get
$$ r_1 + r_2 = \frac{ \frac{466}{81} + 2\left(\frac{-1}{3}\right) }{ \frac{-1}{3} } $$
Simplify:
$$ \frac{-1}{3} = \frac{-27}{81} $$
$$ \frac{466}{81} - \frac{54}{81} = \frac{412}{81} $$
Divide:
$$ r_1 + r_2 = \frac{\frac{412}{81}}{\frac{-1}{3}} = \frac{412}{81}\cdot(-3) = -\frac{1236}{81} = -\frac{412}{27} $$
Step 2: Product of the roots $ (r_1r_2) $
Using
$$ r_1r_2 = \frac{ \alpha^4 + \beta^4 + \alpha\beta + \alpha^3\beta^3 }{ \alpha\beta } $$
and substituting
$$ \alpha^4 + \beta^4 = \frac{466}{81}, \quad \alpha\beta = \frac{-1}{3}, \quad \alpha^3\beta^3 = \left(\frac{-1}{3}\right)^3 = \frac{-1}{27} $$
we get
$$ r_1r_2 = \frac{ \frac{466}{81} + \frac{-1}{3} + \frac{-1}{27} }{ \frac{-1}{3} } $$
Convert to a common denominator:
$$ \frac{-1}{3} = \frac{-27}{81}, \quad \frac{-1}{27} = \frac{-3}{81} $$
Simplify:
$$ \frac{466}{81} - \frac{27}{81} - \frac{3}{81} = \frac{436}{81} $$
Divide:
$$ r_1r_2 = \frac{\frac{436}{81}}{\frac{-1}{3}} = \frac{436}{81}\cdot(-3) = -\frac{1308}{81} = -\frac{436}{27} $$
Step 3: Form the equation
The quadratic equation is
$$ x^2 - (r_1 + r_2)x + r_1r_2 = 0 $$
Substitute:
$$ x^2 - \left( -\frac{412}{27} \right)x - \frac{436}{27} = 0 $$
Simplify:
$$ x^2 + \frac{412}{27}x - \frac{436}{27} = 0 $$
Multiply through by $27$:
$$ 27x^2 + 412x - 436 = 0 $$
Final Answer
The quadratic equation is
$$ \boxed{ 27x^2 + 412x - 436 = 0 } $$
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