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Given, x + y + z = 19, a2 + b2 + c2 = 133 and ab = y2 We know, (a + b + c)2 = (a2 + b2 + c2 ) + 2 × (ab + bc + ac) Or, (19)2 = 133 + 2 × (ab + bc + ac) Or, (ab + bc + ca) = (361 – 133) ÷ 2 Or, (ab + bc + ca) = 114 Or, (ab + bc + b2 ) = 114 Or, b(a + b + c) = 114 Or, b × 19 = 114 Or, b = (114/19) = 6 Therefore, ac = 62 = 36 Therefore, possible values of ‘a’ and ‘c’ = are 4 or 9 Required difference = 9 – 4 = 5
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