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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.
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}.
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.  
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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