How do you solve x^2+8x-2=0x2+8x2=0 using completing the square?

3 Answers
Jun 18, 2018

Solution: x = (-4 +3 sqrt 2) ,x= ( -4 - 3sqrt 2) x=(4+32),x=(432)

Explanation:

x^2+8 x-2=0 or x^2+8 x=2x2+8x2=0orx2+8x=2 or

x^2+8 x +16 =2+16 x2+8x+16=2+16 or

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

x+4 = +- 3sqrt 2 or x = -4+- 3 sqrt 2x+4=±32orx=4±32

Solution: x = (-4 +3 sqrt 2) ,x= ( -4 - 3sqrt 2) x=(4+32),x=(432)[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+sqrt18x=4+18 and x=-4-sqrt18x=418

Explanation:

To complete the square, take the second coefficient (the one next to the xx), 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(8÷2)2=16

Add this number to both sides of the original equation:
(x^2+8x+16) - 2 = 16(x2+8x+16)2=16
(x+4)^2 - 2 = 16(x+4)22=16
(x+4)^2 = 18(x+4)2=18

This gives us 2 possible solutions:
x+4 = +-sqrt18x+4=±18

x=-4+-sqrt18x=4±18

Jun 18, 2018

x=-4+-3sqrt2x=4±32

Explanation:

"add 2 to both sides"add 2 to both sides

x^2+8x=2x2+8x=2

"add "(1/2"coefficient of the x-term")^2" to both sides"add (12coefficient of the x-term)2 to both sides

x^2+2(4)x color(red)(+16)=2color(red)(+16)x2+2(4)x+16=2+16

(x+4)^2=18(x+4)2=18

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

sqrt((x+4)^2)=+-sqrt18larrcolor(blue)"note plus or minus"(x+4)2=±18note plus or minus

x+4=+-sqrt18=+-sqrt(9xx2)=+-3sqrt2x+4=±18=±9×2=±32

"subtract 4 from both sides"subtract 4 from both sides

x=-4+-3sqrt2larrcolor(red)"exact values"x=4±32exact values

x~~-8.24" or "x~~0.24" to 2 dec. places"x8.24 or x0.24 to 2 dec. places