How do you solve #x^2+4x+2=0#?

1 Answer
Jun 30, 2015

Answer:

#x=-2+sqrt(2)#
or
#x=-2-sqrt(2)#

Explanation:

For any equation of the form
#color(white)("XXXX")##ax^2+bx+c=0#
the solutions can be obtained using the quadratic formula:
#color(white)("XXXX")##x= (-b+-sqrt(b^2-4ac))/(2a)#
(this formula shows up often enough that it is worth memorizing)

Substituting the values from #x^2+4x+2 = 0# for #a,b, and c#
#color(white)("XXXX")##x= (-4+-sqrt(4^2-4(1)(2)))/(2(1))#

#color(white)("XXXX")##x = (-4+-sqrt(8))/2#

#color(white)("XXXX")##x=-2+-sqrt(2)#