I. 2x2 – 15x + 27 = 0 II. 2y2 – 17y + 33 = 0

Explanation

I. 2x2 – 15x + 27 = 0 ⇒ 2x2 – 6x – 9x + 27 = 0 ⇒ 2x(x – 3) –9(x – 3) = 0 ⇒ (x – 3) (2x – 9) = 0 ⇒ x = 3, 9/2 II. 2y2 – 17y + 33 = 0 ⇒ 2y2 – 6y –11y + 33 = 0 ⇒ 2y(y – 3) –11(y – 3) = 0 ⇒ (2y – 11)(y – 3) = 0 ⇒ y = 11/2, 3 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 = 3, 9/2 So, roots of second equation = y = 11/2, 3 After comparing we can conclude that relationship cannot be established between x and y. IBPS RRB Scale I 2021 Prelims Memory Based Paper