Question #fee42

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Question #fee42

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Answer 1 · 2016-12-27T17:20:57

$a + b \in {- 1, 0}$ $a + b in {-1, 0}$

Explanation:

$x^{2} - S x + P = 0$ $x^2 - Sx + P = 0$

(1) $S = a + b = - a$ $S = a + b = -a$
(2) $P = a b = b \Rightarrow b \frac{a b - b}{b} = 0 \Rightarrow b (a - 1) = 0 \Rightarrow a = 1 or b = 0$ $P = ab = b \Rightarrow b (ab - b)/b = 0 \Rightarrow b(a - 1) = 0 \Rightarrow a = 1 or b = 0$

if a = 1 then by (1), $1 + b = - 1 \Rightarrow b = - 2$ $1 + b = -1 \Rightarrow b = -2$

if b = 0 then by (1), $a + 0 = - a \Rightarrow 2 a = 0 \Rightarrow a = 0$ $a + 0 = -a \Rightarrow 2a = 0 \Rightarrow a = 0$

$(a, b) = (1, - 2) or (0, 0)$ $(a,b) = (1, -2) or (0,0)$

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Question #fee42

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