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 Thread Tools Display Modes 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) , factor the trinomial into . Therefore, each side is and the perimeter will be . You can use a similar process to find (b). For (c), use the format __________________ 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 ? Thread Tools Show Printable Version Email this Page Display Modes Switch to Linear Mode Hybrid Mode Switch to Threaded Mode Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Forum Rules
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