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

Question

Abigail Allen · Accepted Answer

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

Isaiah Anthony · 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# 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#

Genesis Allen · Answer

#x=-4+-3sqrt2# #"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"#

Eva Adamson · Answer

undefined To solve $x^2 + 8x - 2 = 0$ using completing the square:

1. Move the constant term to the other side:
   $x^2 + 8x = 2$

2. To complete the square, take half of the coefficient of $x$, square it, and add it to both sides of the equation:
   $x^2 + 8x + (8/2)^2 = 2 + (8/2)^2$
   $x^2 + 8x + 16 = 2 + 16$

3. Simplify:
   $x^2 + 8x + 16 = 18$

4. Rewrite the left side as a perfect square:
   $(x + 4)^2 = 18$

5. Take the square root of both sides:
   $x + 4 = \pm \sqrt{18}$

6. Simplify:
   $x = -4 \pm \sqrt{18}$

7. Further simplify if needed.