Find the values of a2+b2and a4+b4 when the a+b=5and ab=5
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Solve the system of equations: ìa + b = 5 í î ab = 6 Solve the first for b: a + b = 5 b = 5 - a Substitute 5 - a for b in the second equation of the system: ab = 6 a(5 - a) = 6 5a - a² = 6 -a² + 5a - 6 = 0 a² - 5a + 6 = 0 (a - 3)(a - 2) = 0 a - 3 = 0; a - 2 = 0 a = 3; a = 2 b = 5 - a b = 5 - a b = 5 - 3 b = 5 - 2 b = 2 b = 3 There are two solutions (a,b) = (3,2) and (a,b) = (2,3) But we are told that a > b so we choose the first and discard the second, so a=3 and b=2 Therefore the value of 4(a² - b²) – a³ + b³ is 4(3² - 2²) – 3³ + 2³ 4(9 - 4) - 27 + 8 4(5) - 19 20 - 19 1
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Solve the system of equations: ìa + b = 5 í î ab = 6 Solve the first for b: a + b = 5 b = 5 - a Substitute 5 - a for b in the second equation of the system: ab = 6 a(5 - a) = 6 5a - a² = 6 -a² + 5a - 6 = 0 a² - 5a + 6 = 0 (a - 3)(a - 2) = 0 a - 3 = 0; a - 2 = 0 a = 3; a = 2 b = 5 - a b = 5 - a b = 5 - 3 b = 5 - 2 b = 2 b = 3 There are two solutions (a,b) = (3,2) and (a,b) = (2,3) But we are told that a > b so we choose the first and discard the second, so a=3 and b=2 Therefore the value of 4(a² - b²) – a³ + b³ is 4(3² - 2²) – 3³ + 2³ 4(9 - 4) - 27 + 8 4(5) - 19 20 - 19 1
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