How do you solve #(x - 2) ^ { 2} + 3= 4#?

2 Answers
Apr 28, 2018

Answer:

#x=3 and x=1#

Explanation:

first expand the quadratic equation
#x^2-4x+4+3=4#
then sum all the unknown and the constant which will give you
#x^2-4x+3=0#
then factorise the equation and u get
#(x-3)(x-1)=0#
therefore #x=3 and x=1#

Apr 28, 2018

Answer:

#x=1" or "x=3#

Explanation:

#"isolate "(x-2)^2" by subtracting 3 from both sides"#

#rArr(x-2)^2=4-3=1#

#color(blue)"take the square root of both sides"#

#sqrt((x-2)^2)=+-sqrt1larrcolor(blue)"note plus or minus"#

#rArrx-2=+-1#

#"add 2 to both sides"#

#rArrx=2+-1#

#rArrx=2-1" or "x=2+1#

#rArrx=1" or "x=3#