Question
In each of these questions, two equations (I) and (II)
are given.You have to solve both the equations and give answer  I. 7x² - 19x + 10 = 0                                        II. 8y² + 2y – 3 = 0Solution
7x² - 19x + 10 = 0 7x² - 14x – 5x + 10 = 0 7x (x - 2) – 5 (x - 2) = 0 x = 2, 5/7 II. 8y² + 2y – 3 = 0 8y² + 6y – 4y – 3 = 0 2y(4y + 3) – 1 (4y + 3) = 0 y = −3/4, 1/2 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 = 2, 5/7 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 = -3/4, ½ After comparing roots of quadratic equation we can conclude that x> y.
I. p2 – 15p + 56 = 0
II. q2 + 2q – 63 = 0
I. 22x² - 97x + 105 = 0
II. 35y² - 61y + 24 = 0
I. 4x² -  15x + 9 = 0
II. 20y² -  23y + 6 = 0
Solve the quadratic equation:
5x² − 13x + 6 = 0
I. 2x2 – 5x - 12 = 0
II. y2 – 11y + 30 = 0
I. 2 x² - 15 + 18 = 0
II. x²- 3 + 2 = 0
...I. 18p²- 21p + 6 = 0   Â
II. 16q² - 24q +9 = 0
I. 99x² + 161 x + 26 = 0
II. 26 y² + 161 y + 99 = 0
Roots of the quadratic equation 2x2 + x – 528 = 0 is
l). p² - 29p + 204 = 0
ll). q² + 4q - 221 = 0