How do you solve # x^2-2x-4=0# by completing the square?

2 Answers
Mar 3, 2018

Answer:

#color(purple)(x = sqrt5 + 1, -sqrt5 + 1 = 3.236, -1.236#

Explanation:

#x^2 - 2x - 4 = 0#

#x^2 -2x = 4#

Add 1 to both sides,

#x^2 - 2x + 1 = 4 + 1 = 5#

#(x-1)^2 = (sqrt5)^2#

#x-1 = +- sqrt5#

#color(purple)(x = sqrt5 + 1, -sqrt5 + 1 = 3.236, -1.236#

Mar 3, 2018

Answer:

#x=1+-sqrt5#

Explanation:

#"using the method of "color(blue)"completing the square"#

#• " the coefficient of the "x^2" term must be 1 which it is"#

#• " add/subtract "(1/2"coefficient of x-term")^2" to"#
#x^2-2x#

#rArrx^2+2(-1)xcolor(red)(+1)color(red)(-1)-4=0#

#rArr(x-1)^2-5=0larrcolor(blue)"add 5 to both sides"#

#rArr(x-1)^2=5#

#color(blue)"Take square root of both sides"#

#rArrx-1=+-sqrt5larrcolor(blue)"note plus or minus"#

#"add 1 to both sides"#

#rArrx=1+-sqrt5larrcolor(red)"exact solutions"#