Transcript ax  bx  c  0, a  0 b  b  4ac x 2a The Discriminant It is sometimes enough to know.

ax  bx  c  0, a  0
2
b  b  4ac
x
2a
2
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.
b  b  4ac
x
2a
2
Example
For the equation 4x2 – x + 1 = 0, determine what
type of number the solutions are and how many
exist.
Solution
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
For the equation 5x2 – 10x + 5 = 0, determine what
type of number the solutions are and how many exist.
Solution
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
For the equation 2x2 + 7x – 3 = 0, determine what
type of number the solutions are and how many
exist.
Solution
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
Find an equation for which 5 and –4/3 are solutions.
Solution
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
Find an equation for which 3i and –3i are solutions.
Solution
x = 3i or x = –3i
x – 3i = 0 or x + 3i = 0
Get 0’s on one side
(x – 3i)(x + 3i) = 0
Using the principle
of zero products
x2 – 3ix + 3ix – 9i2 = 0
x2 + 9 = 0
Multiplying
Combining like terms
Example
Find an equation for which – 4, 0 and 1 are solutions.
Solution