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Your solution is lengthy and complicated. Frist find out a^2 + b^2 using a+b whole square formula. Then find out a-b whole square and then find out a-b and then a and b
5+5 =10 2^3+2^3 2^1+1^3 2+3 (b ➖ 3a+2). ab=10 2^5 (ab ➖ 5ab+2).
Substitute x=√a=5+z, y=√b=5-z, and (5+z)(5-z)=25-z^2=10 so z^2=15, z=±√15, a=(5±√15)^2=40±10√15, and b=(5∓√15)^2=40∓10√15
A Nice Math Olympiad Algebra Problem: √a + √b = 10, √(ab) = 10, a, b =?a, b ≠ 0; (√a - √b)² = (√a + √b)² - 4√(ab) = 10² - 4(10) = 60 = 4(15) = (2√15)²√a - √b = ± 2√15, √a + √b = 10; 2√a = 10 ± 2√15, √a = 5 ± √15, √b = 5 -/+ √15a = (5 ± √15)² = 5² ± 10√15 + 15 = 40 ± 10√15, b = 40 -/+ 10√15Answer check:a = 40 ± 10√15, √a = 5 ± √15; b = 40 -/+ 10√15, √b = 5 -/+ √15√a + √b = 10 = (5 ± √15) + (5 -/+ √15) = 10; Confirmed√(ab) = (5 ± √15)(5 -/+ √15)] = 25 - 15 = 10; ConfirmedFinal answer:a = 40 + 10√15, b = 40 - 10√15 or a = 40 - 10√15, b = 40 + 10√15
Your solution is lengthy and complicated. Frist find out a^2 + b^2 using a+b whole square formula. Then find out a-b whole square and then find out a-b and then a and b
5+5 =10 2^3+2^3 2^1+1^3 2+3 (b ➖ 3a+2). ab=10 2^5 (ab ➖ 5ab+2).
Substitute x=√a=5+z, y=√b=5-z, and (5+z)(5-z)=25-z^2=10 so z^2=15, z=±√15, a=(5±√15)^2=40±10√15, and b=(5∓√15)^2=40∓10√15
A Nice Math Olympiad Algebra Problem: √a + √b = 10, √(ab) = 10, a, b =?
a, b ≠ 0; (√a - √b)² = (√a + √b)² - 4√(ab) = 10² - 4(10) = 60 = 4(15) = (2√15)²
√a - √b = ± 2√15, √a + √b = 10; 2√a = 10 ± 2√15, √a = 5 ± √15, √b = 5 -/+ √15
a = (5 ± √15)² = 5² ± 10√15 + 15 = 40 ± 10√15, b = 40 -/+ 10√15
Answer check:
a = 40 ± 10√15, √a = 5 ± √15; b = 40 -/+ 10√15, √b = 5 -/+ √15
√a + √b = 10 = (5 ± √15) + (5 -/+ √15) = 10; Confirmed
√(ab) = (5 ± √15)(5 -/+ √15)] = 25 - 15 = 10; Confirmed
Final answer:
a = 40 + 10√15, b = 40 - 10√15 or a = 40 - 10√15, b = 40 + 10√15