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I. 3x2 – 16x + 21 = 0 => 3x2 – 9x - 7x + 21 = 0 => 3x (x – 3) – 7 (x – 3) = 0 => x = 3, 7/3 II. y2 – 13y + 42 = 0 => y2 – 6y – 7y + 42 = 0 => y (y – 6) – 7(y – 6) = 0 => y = 7, 6 Hence, x < y Alternate Method: if signs of quadratic equation is - ve and +ve respectively then the roots of equation will be +ve and +ve. So, roots of first equation = x = 7/3, 3 So, roots of second equation = y = 7, 6 After comparing we can conclude that x < y.
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