Algebra Period 3 (Friday)

Quadratic Equations: 13-1

But first… Finding the x intercepts of a parabola:
x intercepts = roots = solutions = zeros of a quadratic
You can find these one of two ways:
1) Read them from the graph

2) Set y or f(x) = 0 and then solve

Where the graph crosses the x axis is/are the x intercepts.

(Remember, y = 0 here!)
The x intercepts are the two solutions or roots of the quadratic.
When we factored in Chapter 6 and set each piece equal to zero,
We were finding the x value when y was zero.
That means we were finding these two roots!

You can find these roots (solutions, x intercepts, zeros) by several different methods.
You already know the following 2 methods:

1) Graphing and see where the graph crosses the x axis
2) Factoring, then using the zero products property to solve for x for each factor

ANOTHER METHOD:
Chapter 13-2 Solving Quadratics Using Square Roots -
You did this in Ch 11 for Pythagorean Theorem!
If there is no middle x term, it's easiest to just square root both sides to solve!

BUT THE DIFFERENCE FROM PYTHAGOREAN--NOT LOOKING FOR JUST THE PRINCIPAL SQUARE ROOT ANY MORE.

YOU NEED THE + OR - SYMBOL!!

EXAMPLE:
3x² = 18
divide both sides by 3 and get: x² = 6
square root each side and get x = + or - SQRT of 6

HARDER EXAMPLE:
(x - 5)² = 9
SQRT each side and get: x - 5 = + or - 3
+ 5 to both sides: x = 5 + 3 or x = 5 - 3
So, the 2 roots are x = 8 or x = 2

HARDEST EXAMPLE:
(x + 2)² = 7
SQRT each side and get: x + 2 = + or - SQRT of 7
-2 to both sides: x = -2+ SQRT 7 or x = -2 - SQRT 7