Solution

I. 12x²+ 22x + 8 = 0 => 12x²+ 16x + 6x + 8 = 0 => 4x(3x + 4) + 2(3x + 4) = 0 => (3x + 4) (4x + 2) = 0 => x = -4/3, -1/2 II. 4y²- y − 3 = 0 => 4y²- 4y + 3y - 3 = 0 => 4y(y – 1) + 3(y – 1) = 0 => (y – 1) (4y + 3) = 0 => y = 1, y = -3/4 Hence, relation cannot be established between x and 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 = -4/3, -1/2 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 larger root) So, roots of second equation = y = 1, -3/4 After comparing we can conclude that relationship cannot be established between x and y.