Here you will see students as young as 4 and 5 years old doing algebra and "advanced" math, without ever knowing it's supposed to be hard.
You are invited to learn how to use this method...

Wednesday, December 22, 2010

Directed Discovery With Higher Powers

How would you solve this equation: 4x4 - 2x2 - 4 = 2

I saw this on a math Q and A site.

Hero zero and no fun get back to one.

4x4 - 2x2 - 4 - 2 = 0
4x4 - 2x2 - 6 = 0

You have to know how to factor polynomials.

2(2x4 - x2 - 3) = 0
2(2x2 - 3)(x2 + 1) = 0

or just

(2x4 - x2 - 3) = 0
(2x2 - 3)(x2 + 1) = 0

because hero zero is impervious to division by zero; then just set each factor to zero and solve for x:

2x2 - 3 = 0 and x2 + 1 = 0
2x2 = 3 and x2 = -1

because if one factor is zero the whole thing is zero

x2 = 3/2 and x = ±√-1 [√-1 gives nonreal solution, an imaginary solution, which is ±i ]

x = ±√(3/2)

the real values of x are √(3/2) and -√(3/2)
the imaginary or non real values are +i and -i

I put this here without the blocks or pictures of them because it shows where you can go once you get a feel for the higher powers. Also as a clue or some directed discovery to those who have done some study on this blog and website or have started using base ten blocks but don't have hours and hours of training under their belt yet. Note that

2x4 - x2 - 3

is comprised of a square some rectangles and some smaller squares...and that we could use a big red square the blue x's and the green units to model

2x4 - x2 - 3 instead of basic operations pieces which would be the big blue squares the red strips and the units

really it's the same problem as

2x2 - x - 3 isn't it?
The x's are just raised by a factor of 2...and with our imaginations we can name the pieces anything we want...and you can see if a 12 year old can play with problems like 2x2 - x - 3 and look at them as puzzles not things to be feared and in fact as confidence grows this stuff is fun and challenging...and easily solved given time.

You can see how to do the negatives for a buck on my website. And shortly you will find a video showing this very problem on the advanced algebra page. To get to the advanced algebra page you need a password and currently passwords are a buck. I am told I need to raise the price to 20 bucks because nobody thinks there's any value there for just a dollar. Get it? Any value!!! I assure you there are lots of values there...

BTW this is post 100!!!

250 to go and search engines will start taking this blog seriously...

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