Let $\alpha \in\mathbb{R}$ and let $\alpha,\beta$ be the roots of the equation ${x^2} + {60^{{1 \over 4}}}x + a = 0$. If ${\alpha ^4} + {\beta ^4} = - 30$, then the product of all possible values of $a$ is ____________.
Answer (integer)
45
Solution
$x^{2}+60^{\frac{1}{4}} x+a=0$
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$\therefore \alpha+\beta=-60^{\frac{1}{4}}, \alpha \beta=a$
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Now $\alpha^{4}+\beta^{4}=-30$
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$\Rightarrow\left(\alpha^{2}+\beta^{2}\right)^{2}-2 a^{2}=-30$
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$\Rightarrow\left[(\alpha+\beta)^{2}-2 a\right]^{2}-2 a^{2}=-30$
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$\Rightarrow\left(60^{\frac{1}{2}}-2 a\right)^{2}-2 a^{2}=-30$
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$\Rightarrow 60+4 a^{2}-4 \cdot 60^{\frac{1}{2}} a-2 a^{2}+30=0$
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$\Rightarrow 2 a^{2}-8 \sqrt{15} a+90=0$
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Product of value of $a=45$
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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