I. x2 + 91 = 20x II. 10y2 -

Solution

I. x² + 91 = 20x => x² - 20x + 91 = 0 => x² - 13x - 7x + 91 = 0 =>x(x - 13) - 7(x - 13) = 0 => (x - 13) (x - 7) = 0 => x = 13, 7 II. 10y² - 29y + 21 = 0 => 10y² - 14y - 15y + 21 = 0 => 2y(5y - 7) - 3(5y - 7) = 0 => (5y - 7) (2y - 3) = 0 => y = 7/5, 3/2 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 = 13, 7 So, roots of second equation = y = 7/5, 3/2 After comparing we can conclude that x > y.