I. x2 – 7x + 12 = 0 II.

Solution

I. x² – 7x + 12 = 0 => x² – 4x – 3x + 12 = 0 => x(x – 4) – 3(x – 4) = 0 => (x – 4) (x – 3) = 0 => x = 4, 3 II. y² – 7y + 10 = 0 => y² – 5y – 2y + 10 = 0 => y(y – 5) – 2(y – 5) = 0 => (y – 5) (y – 2) = 0 => y = 5, 2 Hence, relation cannot be established. 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 = 4, 3 So, roots of second equation = y = 5, 2 After comparing we can conclude that relationship cannot be established between x and y.