November 12, 2010

Discriminant

On the 12^th of November, the first thing we did was get into our groups for the “Principia Mathematica Project.” Mr. Piatek mentioned that Project draft is due on November 19^th. Each group were assigned to do a set of questions on the booklet which are found on pages 19 to 21.

Group 1

Number of Solutions: 2

Group 1 had 2 x-intercepts, therefore there are 2 real number solutions.

Group 2

Number of Solutions: 0

Group 2 had no x-intercept, therefore there are 0 real number solutions.

Group 3

Number of Solutions: 1

Group 3 had 1 x-intercept, therefore 1 real number solution.

Using the formula b²-4ac you can determined the discriminant;

Whenever the discriminant is positive, there will be 2 solutions.

x²-4=0

a= 1

b= 0

c= -4

b²- 4ac
(0)²- 4(1)(4)

discriminant= 16

Whenever the discriminant is negative, there will be 0 solutions.

x²-x+2=0

a= 1

b= -1

c= 2

 b²-4ac

(-1)²- 4(1)(2)

discriminant= -7

Whenever the discriminant is zero, there will be 1 solution.

x²-6x+9=0

a= 1

b= -6

c= 9

 b²-4ac

(-6)²- 4(1)(9)

discriminant= 0

Example 1A on page 25.

2x²+3x-10=0

a= 2

b= 3

c= -10

 b²-4ac

(3)²- 4(2)(10) = 89

Since the discriminant is 8, there are 2 solutions.

b²-4ac>0 

2 distinct, 2 real number roots.

Example 2A on page 26.

two distinct real roots?

x²+10x+k=0

a= 1

b= 10

c= k

 b²-4ac>0

(10)²- 4(1)(k)>0

100- 4k >0

-4k> -100

 - 4      -4

k>25

Homework[s]: Discriminant Worksheet, Example 1 and 2 on pages 25 and 26 in the booklet.