Question #c2658

1 Answer
Jan 27, 2018

x = (1+sqrt5), (1-sqrt5)

Explanation:

Given, x^2-2x=4
rArr x^2-2x-4=0
rArr x^2-2x+1-5=0
rArr (x-1)^2-(sqrt5)^2=0
rArr [(x-1)+sqrt5][(x-1)-sqrt5]=0
rArr [(x-1)+sqrt5]=0,[(x-1)-sqrt5]=0
rArr (x-1+sqrt5)=0, (x-1-sqrt5)=0
rArr x = (1-sqrt5), (1+sqrt5)