What is the discriminant of #3x^2 + 6x + 5# and what does that mean?

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Answer 1 · 2015-07-27T12:28:05

For this quadratic, #Delta = -24#, which means that the equation has no real solution, but that it does have two distinct complex ones.

For a quadratic equation written in general form

#ax^2 + bx + c = 0#,

the discriminant is defined as

#Delta = b^2 - 4 * a * c#

In your case, the quadratic looks like this

#3x^2 + 6x +5 = 0#,

which means that you have

#{(a=3), (b=6), (c=5) :}#

The discriminant will thus be equal to

#Delta = 6^2 - 4 * 3 * 5#

#Delta = 36 - 60 = color(green)(-24)#

When #Delta<0#, the equation has no real solutions. It does have two distinct complex solutions derived from the general form

#x_(1,2) = (-b +- sqrt(Delta))/(2a)#

which in this case becomes

#x_(1,2) = (-b +- isqrt(-Delta))/(2a)#, when #Delta<0#.

In your case, these two solutions are

#x_(1,2) = (-6 +- sqrt(-24))/(2 * 3)#

#x_(1,2) = (-6 +- isqrt(24))/6 = (-6 +- 2isqrt(6))/6 = {(x_1 = (-3 - isqrt(6))/3), (x_2 = (-3 + isqrt(6))/3) :}#