An algebra problem by Achal Jain

Algebra Level 3

Let $f(x)= x^{5}+x+1$ , whose roots are $\alpha_1$ , $\alpha_2$ , $\alpha_3$ , $\alpha_4$ , and $\alpha_5$ and let $p(x)= x^{2}-2$ . If the value of $p(\alpha_1)p(\alpha_2)p(\alpha_3)p(\alpha_4)p(\alpha_5) = \lambda$ , what is $\sqrt{|\lambda|}$ ?

Notation: $| \cdot |$ denotes the absolute value function .

8 7 5 6

1 solution

Chew-Seong Cheong
Sep 11, 2017

$\begin{aligned} \lambda & = \prod_{k=1}^5 p(a_k) \\ & = \prod_{k=1}^5 \left(a_k^2 - 2\right) \\ & = \prod_{k=1}^5 \left(a_k - \sqrt 2\right)\left(a_k + \sqrt 2\right) \\ & = \prod_{k=1}^5 \left(\sqrt 2 - a_k \right)\left(-\sqrt 2 - a_k \right) & \small \color{#3D99F6} \text{Note that }f(x) = \prod_{k=1}^5 \left(x - a_k \right) \\ & = f\left(\sqrt 2\right) f\left(-\sqrt 2\right) \\ & = \left(4\sqrt 2+\sqrt 2 +1\right) \left(-4\sqrt 2-\sqrt 2 +1\right) \\ & = \left(5\sqrt 2+1\right) \left(-5\sqrt 2 +1\right) \\ & = -50+1 = -49\end{aligned}$

$\implies \sqrt{|\lambda|} = \sqrt{|-49|} = \boxed{7}$

An algebra problem by Achal Jain

1 solution

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