HW help plz - XP Math - Forums

 XP Math - Forums HW help plz

 12-10-2007 #1 soulsoundUK Guest   Posts: n/a HW help plz 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 Last edited by soulsoundUK; 12-10-2007 at 07:18 PM..
 12-10-2007 #2 Mr. Hui     Join Date: Mar 2005 Posts: 10,609 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 $(7a + ?)(7a + ?)$ __________________ Do Math and you can do Anything! Last edited by Mr. Hui; 12-10-2007 at 11:27 PM..
 12-13-2007 #3 MyriamK Guest   Posts: n/a Watch your formulas 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.
 12-22-2007 #4 Temperal Guest   Posts: n/a Why can't we work in a field of characteristic two and have $(x+y)^2=x^2+y^2$?

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