Question
In each of these questions, two equations (I) and (II)
are given.You have to solve both the equations and give answer  I. x2 – 26x + 168 = 0   II. 2y2 - 19y – 60 = 0Solution
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 = -16/3, 1 if signs of quadratic equation is -ve and -ve respectively then the roots of equation will be +ve and -ve. So, roots of second equation = y = 12, -5/2 After comparing roots of quadratice eqution we can conclude that x ≥ y.
 If 4x = 40, 3y = 33, what is the value of 6x + 4y?
I. 2y² - 3y – 14 = 0
II. 3x² - 7x + 4 = 0
If a quadratic polynomial y = ax2 + bx + c intersects x axis at a and β, then
I. 7x + 8y = 36
II. 3x + 4y = 14
I. 4p² + 17p + 15 = 0
II. 3q² + 19q + 28 = 0
The equation x2 – px – 60 = 0, has two roots ‘a’ and ‘b’ such that (a – b) = 17 and p > 0. If a series starts with ‘p’ such...
I. x2 + 16x + 63 = 0
II. y2 + 2y - 15 = 0
I. 40 x² - 93 x + 54 = 0
II. 30 y² - 61 y + 30 = 0
I. x + 1 = 3√ 9261
II. y + 1 = √ 324
I. 35x² - 24x – 35 = 0
II. 72y² - 145y + 72 = 0
