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Question Number 19247 by 99 last updated on 07/Aug/17

Commented by NEC last updated on 08/Aug/17

$p l e a s e t y p e i t s o w e c a n s e e i t$ $c l e a r l y .$
Commented by Tinkutara last updated on 08/Aug/17

$α, β, γ and δ are the roots of$ $x^{4} + 4 x^{3} - 6 x^{2} + 7 x - 9 = 0 .$ $Then find the value of$ $(1 + α^{2}) (1 + β^{2}) (1 + γ^{2}) (1 + δ^{2})$
	Answered by Tinkutara last updated on 10/Aug/17

	$Since α, β, γ, δ are the roots of the$ $given equation, we have$ $x^{4} + 4 x^{3} - 6 x^{2} + 7 x - 9 =$ $(x - α) (x - β) (x - γ) (x - δ)$ $Putting x = i and then x = - i, we get$ $1 - 4 i + 6 + 7 i - 9 =$ $(i - α) (i - β) (i - γ) (i - δ)$ $and 1 + 4 i + 6 - 7 i - 9 =$ $(- i - α) (- i - β) (- i - γ) (- i - δ)$ $Multiplying these two equations, we get$ $(- 2 + 3 i) (- 2 - 3 i) =$ $(1 + α^{2}) (1 + β^{2}) (1 + γ^{2}) (1 + δ^{2})$ $or$ $13 = (1 + α^{2}) (1 + β^{2}) (1 + γ^{2}) (1 + δ^{2})$
	Commented by ajfour last updated on 10/Aug/17

	$how nice!$
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