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(A) Given the quadratic equation $x^2+60^{\frac{1}{4}} x+a=0$,we have the sum of roots $\alpha+\beta = -60^{\frac{1}{4}}$ and the product of roots $\a

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$45$

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(A) Given the quadratic equation $x^2+60^{\frac{1}{4}} x+a=0$,we have the sum of roots $\alpha+\beta = -60^{\frac{1}{4}}$ and the product of roots $\alpha \beta = a$.We are given $\alpha^4+\beta^4 = -30$.Using the identity $\alpha^4+\beta^4 = (\alpha^2+\beta^2)^2 - 2(\alpha \beta)^2$,we substitute the values:$(\alpha^2+\beta^2)^2 - 2a^2 = -30$.Since $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha \beta = (-60^{\frac{1}{4}})^2 - 2a = 60^{\frac{1}{2}} - 2a$,we have:$(60^{\frac{1}{2}} - 2a)^2 - 2a^2 = -30$.Expanding the square:$60 + 4a^2 - 4a(60^{\frac{1}{2}}) - 2a^2 = -30$.Simplifying the equation:$2a^2 - 4\sqrt{60}a + 90 = 0$.This is a quadratic equation in $a$. The product of the roots $a_1 a_2$ is given by the constant term divided by the leading coefficient:$	ext{Product} = \frac{90}{2} = 45$.

Let $a \in \mathbb{R}$ 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 $......$

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