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At a Glance - How Many Roots?

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 = x2x + 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


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