How do you solve #-x^2+2x-1=0#?

2 Answers
May 29, 2018

Answer:

x=1

Explanation:

IT IS EASY!!
#-x^2+2x-1=0# { * COMPARING WITH
#ax^2+bx+c#
we see
#a=-1,b=2,c=-1#
by splitting the middle term....
WE GET,
PRODUCT=#ca = -1*-1=1#
SUM = # 2#
*
therefore
#-x^2+x+x-1=0#
#-x(x-1)+(x-1)=0#
#(-x+1)(x-1)=0#
hence
x=1................................. hope u get it
* CHEERS !!:D *

May 29, 2018

Answer:

#x=1#.

Explanation:

This equation can be solved by factorization.

#-x^2 + 2x - 1 = 0#

#-1(x^2 -2x + 1) = 0#

#x^2 -2x + 1 = 0#

#x^2 - x - x + 1 =0#

#x( x-1 ) -1( x-1 ) =0#

#( x-1 )( x-1 ) = 0#

Now,
#x-1 = 0#

Therefore, #x=1#.