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

3 Answers
Jun 18, 2018

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

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

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"