JEE MAIN - Mathematics (2023 - 25th January Evening Shift - No. 25)
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
45
Explanation
$x^{2}+60^{\frac{1}{4}} x+a=0$
$$ \therefore \alpha+\beta=-60^{\frac{1}{4}}, \alpha \beta=a $$
Now $\alpha^{4}+\beta^{4}=-30$
$\Rightarrow\left(\alpha^{2}+\beta^{2}\right)^{2}-2 a^{2}=-30$
$\Rightarrow\left[(\alpha+\beta)^{2}-2 a\right]^{2}-2 a^{2}=-30$
$\Rightarrow\left(60^{\frac{1}{2}}-2 a\right)^{2}-2 a^{2}=-30$
$\Rightarrow 60+4 a^{2}-4 \cdot 60^{\frac{1}{2}} a-2 a^{2}+30=0$
$\Rightarrow 2 a^{2}-8 \sqrt{15} a+90=0$
Product of value of $a=45$
$$ \therefore \alpha+\beta=-60^{\frac{1}{4}}, \alpha \beta=a $$
Now $\alpha^{4}+\beta^{4}=-30$
$\Rightarrow\left(\alpha^{2}+\beta^{2}\right)^{2}-2 a^{2}=-30$
$\Rightarrow\left[(\alpha+\beta)^{2}-2 a\right]^{2}-2 a^{2}=-30$
$\Rightarrow\left(60^{\frac{1}{2}}-2 a\right)^{2}-2 a^{2}=-30$
$\Rightarrow 60+4 a^{2}-4 \cdot 60^{\frac{1}{2}} a-2 a^{2}+30=0$
$\Rightarrow 2 a^{2}-8 \sqrt{15} a+90=0$
Product of value of $a=45$
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