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

2 Answers
Jun 28, 2017

Answer:

#x=-4+-sqrt14#

Explanation:

#"express as " x^2+8x=-2#

#"to "color(blue)"complete the square"#

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

#"that is add " (8/2)^2=16" to both sides"#

#rArrx^2+8xcolor(red)(+16)=-2color(red)(+16)#

#rArr(x+4)^2=14#

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

#sqrt((x+4)^2)=+-sqrt14larr" note plus or minus"#

#rArrx+4=+-sqrt14#

#"subtract 4 from both sides"#

#xcancel(+4)cancel(-4)=+-sqrt14-4#

#rArrx=-4+-sqrt14#

Jun 28, 2017

Move +2 to the right side of the equation.
#x^2 + 8x = -2#

Then halve the coefficient of x.
#x^2 + (8/2)x = -2#

Then square that same coefficient.
#x^2 + (8/2)^2x = -2#

Since #(8/2)^2# = #16# we can put that number into the equation.
So, #x^2 + 8x + 16 = -2#

When you find the number to complete the square you must add it to both sides of the equation.
So, #x^2 + 8x + 16 = -2 +16#
= #x^2 + 8x + 16 = 14#

Then factorise #x^2 + 8x + 16# = #(x+4)(x+4)#

Therefore, the answer is: #(x+4)(x+4)=14#