Algebra Period 3 (Friday)

How about one more video..
let me know if you find one you think everyone in our class would like


Solving Rational Expressions: 13-5

YOU NEED TO GET RID OF ALL DENOMINATORS!
After you do that, you may end up with a Quadratic that you can solve using any of your methods.

Easy one first

(a+1)/2 = 1/a
You could find the LCM of the denominators which would be 2a but what about using cross products—as you did with proportions—last year.
then the above becomes
a(a + 1) = 2(1)
0r
a² + a = 2
a² + a - 2= 0 using ZERO products property
Factor
(a + 2)(a-1) = 0
so a = -2 and a= 1


When you MUST Find the LCM of the denominators:
MULTIPLY EACH & EVERY TERM ON BOTH SIDES BY THE LCM!
You should end up with NO DENOMINATORS!
(Otherwise, you don't have the right LCM!)

EXAMPLE:
4/x – 4/(x+2) = 1
In this case the LCM is discovered just by multiplying both of the existing denominators. It is as if they are relatively prime!!
The LCM is x(x + 2)
Multiply each term on BOTH sides by x(x + 2)
[x(x +2)4]/x – [4 x(x +2)]/(x+2) = 1 [x(x +2)]

Simplify the denominators with the numerators:
(x + 2) 4 - x(4) = x(x + 2)
SIMPLIFY more:
4x + 8 - 4x = x² + 2x

USE THE ZERO PRODUCTS PROPERTY:
x² + 2x - 8 = 0
FACTOR OR QUADRATIC FORMULA:
(x + 4)(x - 2) = 0

x = -4 and x = 2


Solving Radical Equations 13-6
THIS IS A REVIEW OF CHAPTER 11

You square both sides to get rid of the radical sign
For these problems, you'll get a quadratic on one side after you square
BE SURE TO CHECK BOTH ANSWERS TO MAKE SURE THEY BOTH CHECK!

x – 5 = SQRT (x+ 7)
squaring both sides gives us
(x -5)² = [SQRT(x + 7)]²

x² - 10x + 25 = x + 7
x²- 11x + 18 = 0
(x – 9)(x -2) = 0
or x = 9 and x = 2

BUT when you check you discover
x -5 = SQRT ( x + 7)
9 – 5 =?= SQRT (9 + 7)
4 = 4
BUT

x -5 = SQRT ( x + 7)
2 – 5 =?= SQRT (2 + 7)
-3 DOES NOT EQUAL 3

so only solution is x = 9

Solve
[SQRT ( 27 - 3x)] + 3 = x
move the 3 to the other side using transformations

[SQRT ( 27 - 3x)] = x -3
NOW squaring both sides

[SQRT ( 27 - 3x)]² = (x -3)²
27 – 3x = x² - 6x + 9
move everything to the right side – to use the ZERO PRODUCTS PROPERTY

0 = x² - 3x – 18
0 = ( x – 6) (x +3)
so x = 6 or x = -3

WE MUST CHECK again:
Start with the ORIGINAL equation
[SQRT ( 27 - 3x)] + 3 = x

Plug in for x = 6
[SQRT ( 27 – 3(6))] + 3 =? = 6
[SQRT ( 27 - 18)] + 3 =? = 6
[SQRT ( 9)] + 3 =?= 6
3 + 3 = 6 YES

But we discover

[SQRT ( 27 - 3x)] + 3 = x
Plug in for x = -3
[SQRT ( 27 – 3(-3))] + 3 =?= -3
[SQRT ( 27 +9)] + 3 =?= -3
[SQRT ( 36)] + 3 =?= -3
6 + 3 DOES NOT EQUAL -3
so only x = 6 is the solution