Quadratic Equation Question 5
Question 5 - 25 January - Shift 2
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
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Answer: 45
Solution:
Formula: Sum and Product of Roots
$\begin{aligned} & x^2+60^{1 / 4} \mathrm{x}+\mathrm{a}=0 \Rightarrow \alpha+\beta=-60^{1 / 4}, \alpha \beta=\mathrm{a} \ & \alpha^2+\beta^2+2 \alpha \beta=\sqrt{60} \ & \alpha^2+\beta^2=\sqrt{60}-2 \mathrm{a} \ & \alpha^4+\beta^4+2 \alpha^2 \beta^2=60+4 a^2-4 \sqrt{60} \mathrm{a} \ & \Rightarrow-30+2 \mathrm{a}^2=60+4 \mathrm{a}^2-4 \sqrt{60} \mathrm{a} \ & \Rightarrow 2 \mathrm{a}^2-4 \sqrt{60} \mathrm{a}+90=0 \ & \text { Product of rots }=\frac{90}{ 2}=45 \end{aligned}$
