Question

Let \(a \in R\) and let \(\alpha, \beta\) be the roots of the equation \(x^2+60^{\frac{1}{4}} x+a=0\). If \(\alpha^4+\beta^4=-30\), then the product of all possible values of \(a\) is_____

Answer 1

Correct Answer: 45

From the quadratic equation:

\( \alpha + \beta = -60^{\frac{1}{4}}, \quad \alpha \beta = a. \)

Using the condition \(\alpha^4 + \beta^4 = -30\):

\( (\alpha^2 + \beta^2)^2 - 2(\alpha \beta)^2 = -30. \)

Substitute \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta\):

\( ((-60^{\frac{1}{4}})^2 - 2a)^2 - 2a^2 = -30. \)

Simplify and solve for \(a\), leading to:

\( 2a^2 - 4 \cdot 60^{\frac{1}{2}}a + 90 = 0. \)

The product of all possible values of \(a\) is:

\( \boxed{45}. \)

Answer 2

The correct answer is 45.

 

 

 

 

$\alpha +\beta =-60^{\frac{1}{4}}\&\alpha \beta =a$
Given ${\alpha }^4+{\beta }^4=-30$
$\Rightarrow {\left({\alpha }^2+{\beta }^2\right)}^2-2{\alpha }^2{\beta }^2=-30$
$\Rightarrow {\left\{(\alpha +\beta {)}^2-2\alpha \beta \right\}}^2-2a^2=-30$
$\Rightarrow {\left\{60^{\frac{1}{2}}-2a\right\}}^2-2a^2=-30$
$\Rightarrow 60+4a^2-4a\times 60^{\frac{1}{2}}-2a^2=-30$
$\Rightarrow 2a^2-4.60^{\frac{1}{2}}a+90=0$
Product $=\frac{90}{2}=45$