When you solve for the roots of a quadratic equation, there are several possible outcomes.
- You can have two real number solutions. If you set x equal to either solution, the result with be zero both times.
- There can be just one real number solution.
- The equation can have two complex number solutions. There are no real number solutions.
Don't worry; there's an easy way to find out how many solutions there are before you even start using the formula. Just take a peek at the b2 – 4ac part of the quadratic formula. That little chunk is called the discriminant, and it's the keystone species of our little quadratic ecosystem. Without it, the whole thing falls apart.
- If b2 – 4ac is positive, then there are two real number solutions.
- If b2 – 4ac = 0, then there's only one real number solution.
- If b2 – 4ac is negative, then there are two complex number solutions.
This all comes directly from the quadratic formula. If the discriminant is positive, then you have , which leads to two real number answers. If it's negative, you have , which gives two complex results. And if b2 – 4ac is 0, then you have , so you have only one solution.
Sample Problem
How many roots does x2 – 3 = 0 have?
To use the discriminant, we first note that a = 1, b = 0, and c = -3.
b2 – 4ac = (0)2 – 4(1)(-3) = 12
So we have two real roots. Hah! Too easy.
Okay, How About This?
How many roots does 2x2 + 8x + 8 = 0 have?
Hey now, stop it with that lip, Subheading. Why not just say "Sample Problem" like you usually do? Anyway, the discriminant for this equation is
b2 – 4ac = (8)2 – 4(2)(8) = 64 – 64 = 0
That means we have one real number root for this equation.
How Do You Like This One, Then?
How many roots does 0.7731x2 – 2.3812x + 4.1111 = 0 have?
Now that's just being mean—but we can still do it. Just let us find our calculator real quick.
b2 – 4ac = (-2.3812)2 – 4(0.7731)(4.1111) ≈ 5.6701 – 12.7132 = -7.0431
That's negative, so there are two complex roots for this equation. Also, the calculator was in the Shmoop massage room, next to a pile of Algebra textbooks. In case you were wondering.
What Was It Doing There?
We may have been multitasking at the time. We're pretty busy, you know.
Example 1
How many real number or complex number roots does the following equation have? y = x2 + 4x + 1 |
Example 2
How many real number or complex number roots does the following equation have? y = x2 – x + 3 |
Example 3
How many real number or complex number roots does the following equation have? y = -2x2 – 4x + 1 |
Example 4
How many real number or complex number roots does the following equation have? y = -9x2 + 6x – 1 |
Exercise 1
How many real number or complex number roots does the following equation have?
y = 3x2 + 3x + 3
Exercise 2
How many real number or complex number roots does the following equation have?
y = x2 + x + 1
Exercise 3
How many real number or complex number roots does the following equation have?
y = -x2 – 5x + 4
Exercise 4
How many real number or complex number roots does the following equation have?
y = -3x2 + 6x – 3
Exercise 5
How many real number or complex number roots does the following equation have?
y = x2 – 2x – 2