How do you solve $x^2+2x-3<0$ ?

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Answer 1 · 2016-10-23T19:05:52

$x^2+2x-3<0$ for $-3 < x < 1$

$x^2+2x-3<0$
$rArr(x+3)(x-1)<0$

$color(blue)(x+3<0)rArrcolor(blue)(x<-3)$
$color(blue)(x+3>0)rArrcolor(blue)(x> -3)$

$color(purple)(x-1<0)rArrcolor(purple)(x<1)$
$color(purple)(x-1>0)rArrcolor(purple)(x>1)$

Then,
$x+3 < 0 and x-1 < 0 for -oo< x< -3$
then, $(x+3)(x-1)>0$ $color(red)(REJECTED)$

$x+3 > 0 and x-1 < 0 f or -3< x< 1$
then, $(x+3)(x-1) < 0$ $color(red)(AC CEPTED)$

$x+3 > 0 and x-1 > 0 for 1< x< +oo$
then, $(x+3)(x-1)>0$ $color(red)(REJECTED)$

Therefore,

$x^2+2x-3<0$ for $-3 < x < 1$

How do you solve x^2+2x-3<0x^2+2x-3<0?