I. x2 + 12√2 x + 22 = 0 II.

Solution

I. x² + 12√2 x + 22 = 0 (x + 11√2) (x + √2) = 0 x = - 11√2, -√2  II. y² - 13√2 y – 28 = 0 (y - 14√2) (y + √2) = 0 y = - √2, 14√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 = -11√2, -√2 if signs of quadratic equation is -ve and -ve respectively then the roots of equation will be +ve and -ve. (note: -ve sign will come in smaller root) So, roots of second equation = y = - √2, 14√2 After comparing roots of quadratic equation we can conclude that x ≤ y.