Question
I. 3x² - 22 x + 40 = 0  II. 4y² + 22y +
24 = 0Â Â Â In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer.Solution
I. 3x² - 22 x + 40 = 0   3x² - 12x - 10x + 40 = 0   3x (x – 4) – 10 (x – 4) (3x – 10 ) (x – 4) x = 10/3, 4 II. 4y² + 22y + 24 = 0  4y² + 16 y + 6 y + 24 = 0   4y (y + 4) + 6( y + 4) (4y + 6) (y + 4) x = - 4 , - 3/2 Hence, x > y.
I. x2Â - 9x - 52 = 0
II. y2Â - 16y + 63 = 0
In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between 'p' and 'q' and choose...
Solve the quadratic equations and determine the relation between x and y:
Equation 1: x² - 42x + 392 = 0
Equation 2: y² - 46y + 480 = 0
I. 63x2 + 148x + 77 = 0
II. 21y2 + 89y + 88 = 0
(i) 2x² – 9x + 10 = 0
(ii) 4y² – 12y + 9 = 0
I. 2x² - 9x + 10 = 0
II. 3y² + 11y + 6 = 0
In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between 'p' and 'q' and choose...
I. 22x² - 97x + 105 = 0
II. 35y² - 61y + 24 = 0
I. 2x2 + 13x + 21 = 0
II. 3y2 + 34y + 63 = 0
Equation 1: x² + 16x + 63 = 0
Equation 2: y² + 10y + 21 = 0
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