What is the discriminant of #m^2 + m + 1 = 0# and what does that mean?

2 Answers
Jul 21, 2015

Answer:

The discriminant #Delta# of #m^2+m+1 = 0# is #-3#.

So #m^2+m+1 = 0# has no real solutions. It has a conjugate pair of complex solutions.

Explanation:

#m^2+m+1 = 0# is of the form #am^2+bm+c = 0#, with #a=1#, #b=1#, #c=1#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = 1^2 - (4xx1xx1) = -3#

We can conclude that #m^2+m+1 = 0# has no real roots.

The roots of #m^2+m+1 = 0# are given by the quadratic formula:

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

Notice that the discriminant is the part inside the square root. So if #Delta > 0# then the quadratic equation has two distinct real roots. If #Delta = 0# then it has one repeated real root. If #Delta < 0# then it has a pair of distinct complex roots.

In our case:

#m = (-b+-sqrt(Delta))/(2a) = (-1 +-sqrt(-3))/2 = (-1 +-i sqrt(3)) / 2#

The number #(-1+i sqrt(3))/2# is often denoted by the Greek letter #omega#.

It is the primitive cube root of #1# and is important when finding all roots of a general cubic equation.

Notice that #(m-1)(m^2+m+1) = m^3 - 1#

So #omega^3 = 1#

Jul 21, 2015

Answer:

The discriminant of #(m^2+m+1=0)# is #(-3)# which tells us that there are no Real solutions to the equation (a graph of the equation does not cross the m-axis).

Explanation:

Given a quadratic equation (using #m# as the variable) in the form:
#color(white)("XXXX")##am^2 + bm +c = 0#

The solution (in terms of #m#) is given by the quadratic formula:
#color(white)("XXXX")##m = (-b+-sqrt(b^2-4ac))/(2a)#

The discriminant is the portion:
#color(white)("XXXX")##b^2-4ac#

If the discriminant is negative
#color(white)("XXXX")#there can be no real solutions
#color(white)("XXXX")#(since there is no real value which is the square root of a negative number).

For the given example
#color(white)("XXXX")##m^2+m+1 = 0#
the discriminant, #Delta# is
#color(white)("XXXX")##(1)^2 - 4(1)(1) = -3#
and therefore
#color(white)("XXXX")#there are no Real solutions to this quadratic.