How do you solve #x^2+8x-2=0# using completing the square?

3 Answers
Jun 18, 2018

Answer:

Solution: # x = (-4 +3 sqrt 2) ,x= ( -4 - 3sqrt 2) #

Explanation:

# x^2+8 x-2=0 or x^2+8 x=2# or

# x^2+8 x +16 =2+16 # or

#(x+4)^2= 18 or (x+4) = +- sqrt 18# or

#x+4 = +- 3sqrt 2 or x = -4+- 3 sqrt 2#

Solution: # x = (-4 +3 sqrt 2) ,x= ( -4 - 3sqrt 2) #[Ans]

Jun 18, 2018

Answer:

Take the second coefficient, divide it by 2, and square it, to complete the square, and obtain the solutions:
#x=-4+sqrt18# and #x=-4-sqrt18#

Explanation:

To complete the square, take the second coefficient (the one next to the #x#), divide it by 2, and square it. This will give you the number you need to complete the square.
In this case, that would be #(8-:2)^2 = 16#

Add this number to both sides of the original equation:
#(x^2+8x+16) - 2 = 16#
#(x+4)^2 - 2 = 16#
#(x+4)^2 = 18#

This gives us 2 possible solutions:
#x+4 = +-sqrt18#

#x=-4+-sqrt18#

Jun 18, 2018

Answer:

#x=-4+-3sqrt2#

Explanation:

#"add 2 to both sides"#

#x^2+8x=2#

#"add "(1/2"coefficient of the x-term")^2" to both sides"#

#x^2+2(4)x color(red)(+16)=2color(red)(+16)#

#(x+4)^2=18#

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

#sqrt((x+4)^2)=+-sqrt18larrcolor(blue)"note plus or minus"#

#x+4=+-sqrt18=+-sqrt(9xx2)=+-3sqrt2#

#"subtract 4 from both sides"#

#x=-4+-3sqrt2larrcolor(red)"exact values"#

#x~~-8.24" or "x~~0.24" to 2 dec. places"#