March 2019 archive

week 7 – pre calc 11

The discriminant is the part of the quadratic formula under the square root.

The discriminant can be positive, zero, or negative, and this determines how many solutions there are to the given quadratic equation.

A positive discriminant indicates that the quadratic has two distinct real number solutions.

A discriminant of zero indicates that the quadratic has a repeated real number solution.

A negative discriminant indicates that neither of the solutions are real numbers.

 

6×2+10x−1=0

a=6
b=10
c=−1

Week 6- precalc 11

This week in pre calculus 11 we learned how to solve quadratic equations, which at first the concept was a bit confusing but after working in the workbook started to get much easier because I was starting to see the patterns of the equations and expressions.

 

Today we actually learned the third way of solving a quadratic equations and with this way it uses a quadratic formula (in image above) and you can take the a,b, and c out of our equations that equal 0. For me it wasn’t to hard to start to understand where you would be inputting these numbers and started putting them in the right spots without even realizing, for me the hardest part was remembering the formula to use.

 

I learned that it’s better to take the time to look over your steps and do them slower so you avoid mistakes instead of rushing through them and messing up because it is very  hard to verify them.

Week 5- Pre calc 11

This week in PreCalc 11 we revisited the idea of factoring polynomial expressions like we learned last year in grade 10.

For example when we started the practicing of factoring expressions I remembered that for the second number in the trinomial it was the sum of adding the two numbers together and the third number in the trinomial was the sum of the two numbers multiplied.

For example:

x2 + 7x + 10

I could go through the list of factors if 10 and then find 2 that add together to get the sum of 7.

1-10

2-5

you notice that 2+5 equals 10 so then you input those into the expression

(x + 2) (x + 5)

 

We also learned what to do when the expression is written like this

9×2 + 6x +1

(when there’s a number in front of the first x)

so the first step that i would do is to start to fill in the box with the numbers that i have so the 9×2 and the 1.

 

 

 

then you would multiply 9×2 by 1 then find the factors of the sum

1-9

3-3

and you can see that the 3-3 equals to 6 so those will  be our missing variables

 

and then you would find what’s common between each line and those would be your expression numbers

( 3x + 1)( 3x + 1)