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

2 Answers
Jun 25, 2015

Answer:

#x=1#

Explanation:

Given
#color(white)("XXXX")##x^2-2x+1 = 0#
Factor the left side to get
#color(white)("XXXX")##(x-1)(x-1) = 0#
From which it follows that
#color(white)("XXXX")##(x-1) = 0#
and
#color(white)("XXXX")##x=1#

Jun 25, 2015

Answer:

This is a quadratic equation, that you can factorise.

Explanation:

Find a pair of numbers that add up to #-2x# and multiplies to #+1#

Maybe you'll see that that would be #-1and-1#

Now you can factorise:
#=(x-1)*(x-1)=0->x-1=0->x=1#

Check your answer:
#1^2-2*1+1=0#
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