# Quadratic Formula: Negative in root?

1. Dec 7, 2004

### CinderBlockFist

What do you do when Root(b^2-4ac) is negative? because you can't have a negative under a root right?

2. Dec 7, 2004

### Tide

If the discriminant is negative then the quadratic equation has no real valued roots. However, in some applications defining the square root of a negative number is very useful. Such numbers are called "imaginary" or "complex" numbers. If you haven't learned about them yet then it's possible you made a numerical error (I am assuming this is from a school assignment).

3. Dec 7, 2004

### cepheid

Staff Emeritus
The quadratic has no real roots, meaning that there are no real numbers that satisfy the equation ax2 + bx + c = 0. The roots are complex numbers. Have you encountered complex numbers before? In general, a complex number z can be expressed as a combination of a real and an imaginary number: z = a + bi, which makes sense because imaginary numbers are numbers that, when squared, yield a negative real number. They are multiples of the imaginary number i, which is defined as follows:

$$i = \sqrt{-1}$$

However, if you have encountered your negative discriminant in a quadratic whose solution represents some physical situation, then you have made an error, because the solution must be real. (It is a "real" situation after all, "imaginary" numbers have no place.)

Edit: oops...that's exactly what Tide posted...I guess just while I was composing this

4. Dec 7, 2004

### CinderBlockFist

well i was trying to factor and find roots for x^2-2x+4=0, then i get a negative under the root.

5. Dec 7, 2004

### Tide

Then you should get something like this if you factor over the complex numbers.

$$x^2 - 2x + 4 = \left(x - 1 + i \sqrt 3\right) \left(x - 1 - i \sqrt 3\right)$$

If that doesn't mean anything to you then go back and be sure you are doing the right problem!

6. Dec 7, 2004

### shmoe

If you haven't seen complex numbers yet, your question could still be fine. Negative under the root just means there are no real roots and your polynomial cannot be factored (over the real numbers). Not every equation will have solutions, "no real roots" is an acceptible response.

It might be a good idea to see what it means in terms of the graph to have no real roots. To make this easier, you can complete the square $$x^2-2x+4=(x-1)^2+3$$ and see it's just a shift of the plain old $$x^2$$ graph.