What kind of solutions does #m^2 + m + 1 = 0# have?

2 Answers
Alan P.
Jul 8, 2015

#m^2+m+1 = 0#
has two imaginary solutions

Explanation:

If expressed in a standard quadratic form
#color(white)("XXXX")##am^2+bm+c=0#

The discriminant #Delta = b^2-4ac#
indicates the number of roots
#Delta ={(>0 rArr "2 Real roots"),(=0 rArr "1 Real root"), (<0 rArr "2 Imaginary roots"):}#

#b^2 - 4ac = 1^2 - 4(1)(1) = -3 <0#

Meave60
Jul 8, 2015

The solutions include an imaginary number, #sqrt(-3)=sqrt 3i#.

Explanation:

#m^2+m+1=0# is in the form of a quadratic equation #ax^2+bx+c=0#, where #a=1,# #b=1,# #c=1#.

Use the quadratic formula.

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

Substitute the values for #a#, #b#, and #c# into the quadratic formula.

#x=(-1+-sqrt(1^2-4*1*1))/(2*1)# =

#x=(-1+-sqrt(1-4))/2# =

#x=(-1+-sqrt(-3))/2#

#x=(-1+-sqrt3i)/2# =

#x=(-1+sqrt3i)/2#

#x=(-1-sqrt3i)/2#

#x=(-1+sqrt3i)/2,##(-1-sqrt3i)/2#

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