How do you find the discriminant of #-x^2-x=4# and use it to determine if the equation has one, two real or two imaginary roots?

1 Answer
May 2, 2018

Answer:

The equation has no real roots and two imaginary roots.

Explanation:

The "discriminant" that the question is referring to is the discriminant of the quadratic formula, which is #b^2-4ac#. I've highlighted it in the quadratic formula here in red:

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

We know that by the fundamental theorem of algebra that any quadratic has exactly #2# roots which can be real or imaginary.

If the value of the discriminant is less than #0#, the quadratic has no real roots. This is because the negative number would make the square root "unsolvable." According to the theorem above, there must be two roots, so that means that they are both imaginary (since there are no real ones).

If the value of the discriminant is more than #0#, the quadratic has two real roots. This is because of the "plus-or-minus" sign before the square root. Since there are only two roots according to the theorem, that means that there are no imaginary roots in this case.

If the value of the discriminant is exactly #0#, the quadratic has one real root. This is because the square root would be zero, eliminating the use of the "plus-or-minus", leaving just one solution. However, in this case, there isn't also one imaginary solution, because imaginary solutions only ever come in pairs (conjugate pairs).

In our case, first, we would have to get the quadratic to equal zero on one side:

#-x^2-x=4#

#-x^2=4+x#

#0=4+x+x^2#

#0=x^2+x+4#

Now, we plug our #a#, #b#, and #c# values (which are #1#, #1#, and #4#, respectively) into the discriminant and see what we get:

#color(white)=b^2-4ac#

#=(1)^2-4(1)(4)#

#=1-16#

#=-15#

Since this number is less than #0#, the quadratic has no real roots. Therefore, it has two imaginary roots. We can verify that it has no real roots by looking at its graph:

graph{-x^2-x-4 [-18.3, 17.73, -13.97, 4.04]}

Hope this helped!