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4 Answers

Even= 2,4 so on. So it is asking you for two even numbers that are followed by each other. eg. 24, 26 so what are to following even numbers add up to 224.
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First, let try to find some squared integers nearer to 224. 14x14=10x(10+4+4)+4x4=10x18+4x4=180+16=196, 15x15=1x2x100+5x5=200+25=225, or 15x15=10x(10+5+5)+5x5=10x20+25=200+25=225, 16x16=10x(10+6+6)+6x6=10x22+6x6=220+36=256. Judging from the results obtained above, the two integers would be somewhere close to 14, 15 and 16. The ones-digit of 224 is 4. This indicates the ones-digit of the first integer is 4, and that of the second is 6. Because 4x6=24, but the ones-digits of the procucts of other even consecutive integers are 0 or 8. Therefore, the two consecutive even integers will be 14 and 16. Let check the product. 14x16=10x(10+4+6)+4x6=10x20+24=200+24=224. 14 and 16 are correct integers. Since the product of two integers with the same signs is positive. -14 and -16 are also the integers that satisfy the conditions. Therfore, the answers are two sets of consecutive even integers: {14,16}, and {-14,-16}.
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Let (k-1) and (k+1) be consecutive even integers where k is an odd integer. This problem can be written as follows: (k-1)(k+1)=224. Remove the brackets and simplify the equation. k^2-1=224, k^2=225.* Let find the value of k. Since 10^2=100, 20^2=400, k must be an odd integer between 10 and 20, or -10 and -20. And since the ones-digit of 225 is 5, the ones-digit of k must be 5 because only 5 makes the ones-digit 5 when it's squared. Therefore, k will be 15 or -15. Let check the square of 15. 15x15=10x(10+5+5)+5x5=10x20+5x5=200+25=225. k=15 or -15 is correct. When k=15, the consecutive even integers are (15-1)=14 and (15+1)=16. When k=-15, the consecutive even integers are (-15-1)=-16 and (-15+1)=-14. Let check the product. 14x16=10x(10+4+6)+4x6=10x20+4x6=200+24=224. Since the product of two negative integers is positive, therfore, (-14)x(-16)=224. The answers are two sets of consecutive even integers: {14,16} and {-14,-16}. * Instead of solving the square root of 225, try guessing.  
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Take the complete the square method to solve this quadratic equation problem. Let F and (F+2) be the first even integer and the second one next to the first, respectively. This problem can be written as follows: F(F+2)=224. Remove the brackets: F^2+2F=224. Add 1 to both sides to rewrite the left-hand side into a perfect square form: F^2+2F+1=224+1. Factor the left-hand side equation and symplify: (F+1)^2=225.* Let take the square root of each side: F+1=15, and -15. F=14, and -16. If F=14, F+2=16, and 14x16=224.** Both 14 and 16 are correct answers. If F=-16, F+2=-14. Since the product of two negative numbers is positive, (-16)x(-14)=16x14=224. Both -16 and -14 are also correct answers. Therefore, the ansers are two sets of consecutive even integers: {14,16} and {-16,-14}. *15x15=1x2x100+5x5=200+25=225. **14x16=10x(10+4+6)+4x6=200+24=224.
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