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soulsoundUK
12-10-2007, 04:30 PM
Squaring a binomial can be compared to finding the area of a square. Multiply the length of the side by itself. For example:
(a + b)^2 = a^2 + 2ab + b^2 or (a-b)^2 = a^2 + 2ab - b^2

Problem: The areas of four squares are listed below. Find the length of the side. Then find the perimeter.

(a) a^2 + 8a + 16

(b) a^2 -10a + 25

(c) 49a^2 -56a + 16

(d) 36a^4 -12a^2b^3 + b^6

Mr. Hui
12-10-2007, 10:23 PM
For (a) {a^2}+8a+16, factor the trinomial into (a+4)(a+4). Therefore, each side is a+4 and the perimeter will be 4(a+4) = 4a+16.

You can use a similar process to find (b).

For (c), use the format (7a + ?)(7a + ?)

MyriamK
12-12-2007, 11:07 PM
You wrote (a-b)^2 = a^2 + 2ab - b^2 but that is wrong. Was it like that all the time and I did not notice before?
(a-b)^2 = a^2 - 2ab + b^2

On the other hand, I agree that (a + b)^2 = a^2 + 2ab + b^2
I see that as the square of a sum being equal to the sum of two squares plus two rectangles. If I have a square garden with sides measuring a (area = a^2), and I want to make it bigger, I can add two rectangles of length a and width b to the South and East sides (area = ab for each one) and a square of side b (area = b^2) to fill the Southeast corner. I end up with a square garden of side a+b. The area is (a+b)^2, and equals the sums of the areas of the two squares and two rectangles listed above.

Temperal
12-22-2007, 01:51 PM
Why can't we work in a field of characteristic two and have (x+y)^2=x^2+y^2? :p