How do you solve #x^2 + 4x + 1 = 0# using the quadratic formula?

1 Answer
Mar 11, 2018

Answer:

#x=-2+-sqrt3#

Explanation:

#"Given the quadratic equation in standard form"#

#•color(white)(x)ax^2+bx+c=0 color(white)(x);a!=0#

#"We can solve for x using the "color(blue)"quadratic formula"#

#•color(white)(x)x=(-b+-sqrt(b^2-4ac))/(2a)#

#x^2+4x+1=0" is in standard form"#

#"with "a=1,b=4" and "c=1#

#rArrx=(-4+-sqrt(16-4))/2=(-4+-sqrt12)/2#

#color(white)(x)=(-4+-2sqrt3)/2=-2+-sqrt3larrcolor(red)"exact values"#