Question
I. 81x - 117√x + 40 = 0 II. 81y - 225√y + 136
= 0Solution
I. 81x - 117√x + 40 = 0 81x - 72√x - 45√x + 40 = 0 9√x (9√x - 8) -5 (9√x - 8)= 0 (9√x -5) (9√x -8) = 0 x = 25/81, 64/81 II. 81y - 225√y + 136 = 0 81y - 72√y - 153√y + 136 = 0 9√y (9√y -8) - 17 (9√y - 8) = 0 (9√y - 8) (9√y-17) = 0 y = 289/81, 64/81 Hence, x ≤ y
I. 3q² -29q +18 = 0
II. 9p² - 4 = 0
I.8(x+3) +Â 8(-x) =72Â
II. 5(y + 5) + 5(-y) = 150Â
In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y.
I. x
I. 20y² - 13y + 2 = 0
II. 6x² - 25x + 14 = 0
I. 15y2 + 26y + 8 = 0
II. 20x2 + 7x – 6 = 0
I. 8x2 - 2x – 15 = 0
II. 12y2 - 17y – 40 = 0
l). 3p + 2q = 27
ll). 4p - 3q = 2
I. x2 + 28x + 96 = 0
II. y2 + 3y - 70 = 0
I. x + 1 = 3√ 9261
II. y + 1 = √ 324
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 97x² - 436x + 339 = 0
Equation 2: 103y² - 460y + 357 = 0
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