Question

Sandra exclaims that her quadratic with a discriminant of –4 has no real solutions. Sandra then puts down her pencil and refuses to do any more work. Create an equation with a negative discriminant. Then explain to Sandra, in calm and complete sentences, how to find the solutions, even though they are not real. (10 points)

Answer 1

This is very good question.

The solutions if they are not real but be non-real complex roots.
Recall that if the discriminant is <0, the quadratic function is either a positive definite or a negative definite.

One such case would be
\[x^2 + 4x + 5 = (x+2)^2 + 1 > 0 \]
Verify that discriminant it is indeed negative:
\[\Delta = b^2 - 4ac = (4)^2 - 4(1)(5) = -4 < 0\]

Solving for 0 we get
\[(x+2)^2 = -1 \rightarrow x+2 = \pm \sqrt{-1}\]

If we define the sqrt of -1 to be iota (i)
Then we can write
\[x = -2 \pm i\]

In general for a quadratic with
\[ax^2 + bx + c = 0\]
We can write the solutions using the Quadratic formula as
\[x = \frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]
However if the discriminant is negative then we can further write this as
\[x = \frac{ -b \pm \sqrt{-(4ac - b^2)} }{2a } = \frac{ -b \pm \sqrt{i^2(4ac-b^2)} }{ 2a }\]
\[= \frac{ -b \pm i \sqrt{ 4ac - b^2}}{ 2a } = \frac{ -b \pm i \sqrt{-{\Delta} }}{ 2a }\]

Thus every quadratic equation with real coefficients has at least two complex roots.