The Discriminant

It is sometimes enough to know what type of number a solution will be, without actually solving the equation. From the quadratic formula, b2 – 4ac, is known as the discriminant. The discriminant determines what type of number the solutions of a quadratic equation are. The cases are summarized on the next slide.

2 4

2

b b acx

a

2 0, 0ax bx c a

2 4

2

b b acx

a

Example

Solution

For the equation 4x2 – x + 1 = 0, determine what type of number the solutions are and how many exist.

First determine a, b, and c: a = 4, b = –1, and c = 1. Compute the discriminant:

b2 – 4ac = (–1)2 – 4(4)(1) = –15.

Since the discriminant is negative, there are two imaginary-number solutions that are complex conjugates of each other.

Example

Solution

For the equation 5x2 – 10x + 5 = 0, determine what type of number the solutions are and how many exist.

First determine a, b, and c: a = 5, b = –10, and c = 5. Compute the discriminant:

b2 – 4ac = (–10)2 – 4(5)(5) = 0.

There is exactly one solution, and it is rational. This indicates that 5x2 – 10x + 5 = 0 can be solved by factoring.

Example

Solution

For the equation 2x2 + 7x – 3 = 0, determine what type of number the solutions are and how many exist.

First determine a, b, and c: a = 2, b = 7, and c = –3.

Compute the discriminant:

b2 – 4ac = (7)2 – 4(2)(–3) = 73.

The discriminant is a positive number that is not a perfect square. Thus there are two irrational solutions that are conjugates of each other.

Writing Equations from Solutions

We know by the principle of zero products that (x – 1)(x + 4) = 0 has solutions 1 and -4. If we know the solutions of an equation, we can write an equation, using the principle in reverse.

Example

Solution

Find an equation for which 5 and –4/3 are solutions.

x = 5 or x = –4/3

x – 5 = 0 or x + 4/3 = 0

(x – 5)(x + 4/3) = 0

x2 – 5x + 4/3x – 20/3 = 0

3x2 – 11x – 20 = 0

Get 0’s on one side

Using the principle of zero products

Multiplying

Combining like terms and clearing fractions

Example

Solution

Find an equation for which 3i and –3i are solutions.

x = 3i or x = –3i

x – 3i = 0 or x + 3i = 0

(x – 3i)(x + 3i) = 0

x2 – 3ix + 3ix – 9i2 = 0

x2 + 9 = 0

Get 0’s on one side

Using the principle of zero products

Multiplying

Combining like terms

ExampleFind an equation for which – 4, 0 and 1 are solutions.

Solution