Question

X:X€R,X^2+X-2 =0 in roster Form

Answer 1

$\sf \frak \red{Answer:-}$

$\tt{ {x}^{2} + x - 2 = 0}$

$= > \tt{ {x}^{2} + (2 - 1)x - 2 = 0 }$

$= > \tt{ {x}^{2} + 2x - x - 2 = 0 }$

$= > \tt{x(x + 2) - 1(x + 2) = 0}$

$= > \tt{(x + 2)(x - 1) = 0}$

$= > \tt{x = - 2 \: \: or \: \: x = 1}$

$\therefore \rm{Roster \: form=(-2,1)}$

Answer 2

answer. -2,1 plz mark me as brainliest