Question #fee42

Dec 27, 2016

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

Explanation:

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

(1) $S = a + b = - a$
(2) $P = a b = b \setminus R i g h t a r r o w b \frac{a b - b}{b} = 0 \setminus R i g h t a r r o w b \left(a - 1\right) = 0 \setminus R i g h t a r r o w a = 1 \mathmr{and} b = 0$

if a = 1 then by (1), $1 + b = - 1 \setminus R i g h t a r r o w b = - 2$

if b = 0 then by (1), $a + 0 = - a \setminus R i g h t a r r o w 2 a = 0 \setminus R i g h t a r r o w a = 0$

$\left(a , b\right) = \left(1 , - 2\right) \mathmr{and} \left(0 , 0\right)$