Amy's Reflection #29

okay here's some stuff that might to be helpful on the upcoming test...

Completing the Square:

You can use completing the square to solve a quadratic equation when factoring doesn’t work. This method can only work when 1 is the coefficient of x².

For example:

x² + 6x - 2 = 0

* anytime you are solving a quadratic you’re finding x-intercepts


Move the constant term to the right side:
x² + 6x = 2

Take half of the coefficient on the x-term (divide it by two, and keeping the sign), and then square it. Add the squared value to both sides of the equation:

x² + 6x + 9 = -2 + 9

Convert the left-hand side to squared form. Simplify the right-hand side:

(x + 3)² = 7


* the # half of the coefficient goes in the parentheses.


Square-root both sides:

x + 3 = √7


Solve for "x =". Remember to put the "±" on the right side and that it gives you two solutions.

x = -3 ± √7

The two points for this solution are:

(-3 + √7) , (-3 -√7)

Rational Root therom

Example: f(x)= 2x^3 + 3x^2 - 8 + 3

Step 1: find all possible roots..

p: factors of 3: 1, -1, 3, -3
q: factors of 2: 1, -1, 2, -2

*p is the leading constant term & q is the leading coefficient

possible roots are (p/q): 1, -1, 1/2, -1/2, 3, -3, 3/2, -3/2

Step 2: now you can plug all of the possible roots in your calculator to find the roots that work

the zero will be: 1, 1/2, -3
Step 3: use synthetic division to factor all of the roots that work

you should get: (x - 1) (2x^2 + 5x + 3)

Step 4: slove further

(this can be factored...)

= (x - 1) (2x^2 + 5x + 3)

= (x - 1) (2x - 1) (x + 3)

(set x = 0 )

x = 1, 1/2, -3

i hope that this helps..