Question
I. 3y² - 20y + 25 = 0 II. 3x² - 8x + 5
= 0Solution
I. 3y² - 20y + 25 = 0 3y² - 15y - 5y + 25 = 0 3y (y – 5) – 5 (y – 5) = 0 (3y – 5) (y – 5) = 0 y = 5, 5/3 II. 3x² - 8x + 5 = 0 3x² - 5x - 3x + 5 = 0 (x – 1) (3x – 5) = 0 x = 1, 5/3 Hence, x ≤ y
I. 4x2 – 53x – 105 = 0
II. 3y2 – 25y + 48 = 0
Equation 1: x² - 144x + 5184 = 0
Equation 2: y² - 130y + 4225 = 0
I. x2 – 12x + 32 = 0
II. y2 + y - 20 = 0
I. 15y2 + 26y + 8 = 0
II. 20x2 + 7x – 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 x and y.
I. x
Roots of the quadratic equation 2x2 + x – 528 = 0 is
I. p2Â - 53p + 672 = 0
II. q2Â - 27q + 126Â = 0
I. 2x² - 9x + 10 = 0
II. 3y² + 11y + 6 = 0
I. √(74x-250 )– x=15
II. √(3y²-37y+18)+ 2y=18
Equation 1: x² - 45x + 500 = 0
Equation 2: y² - 60y + 600 = 0
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