Solve the following: `(i)|x-2|=1` (ii) `2|x+1|^2-|x+1|=3`

Explanation

(i) |x-2| =1 I,e those points on real number line which are at distance 1 untis from 2 .
As shown in the figure x=1 and x=3 are at distane 1 untis from 2.
As shown in the figure x=1 and x=3 are at distance 1 unit from2 ltbr. Hence , x=1 or x =3 Also |x-2 |=1
`rArr x-2 = pm 1`
`rArr x=1 or x=3`
(ii) `2 |x+1|^2-|x+1|=3`
`rArr " "2|x+1|^2-|x+1|=3`
`rArr " " 2| x+1|^2-3|x+1|+2|x+1|-3=0`
`rArr " "(2|x+1|-3|x+1|+1)=0`
`rArr " "2|x+1| -3 = 0 `
`rArr " "|x+1|=3//2`
`rArr " " x+1 = pm 3//2`
`rArr x=1//2 or x = -5//2`