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If √a²+b²=613 then find a+b |
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Answer» a² + b² = 613 We see if (a,b,613) is a Pythagorean TRIPLE
let b = 613-k a² + (613-k)² = 613² a² + 613² -1226k + k² = 613² a² - 1226k + k² = 0 We notice that sqrt%281226%29=%2235.01142...%22 So the LARGEST square not exceeding 1226 is 35² = 1225 So we WRITE 1226 = 35² + 1 a² - (35²+1)k + k² = 0 a² = (35²+1)k - k² a² = 35²k + k - k² We see that if k=1 the right SIDE BECOMES 352 Therefore k=1, and b = 613-k = 613-1 = 612, and a=35 Therefore (a,b,613) = (35,612,613) is a Pythagorean triple. Thus a+b = 35+612 = 647 |
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