Question
I. 3y2 + 16y + 16 = 0 II.
2x2 + 19x + 45 = 0Solution
I. 3y2 + 16y + 16 = 0 3y2 + 4y + 12y + 16 = 0 y (3 y + 4) + 4(3 y + 4) = 0 (y + 4)(3 y + 4) = 0 y = - 4, -4/3 II. 2x2 + 19x + 45 =0 2x2 + 9x + 10 x + 45 =0 x (2 x + 9) + 5 (2 x + 9) = 0 (x + 5) (2 x + 9) = 0 x = - 5, -9/2 Hence, x > y
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 21x² - 82x + 80 = 0
Equation 2: 23y² - 132y + 85 = 0
Equation 1: x² - 200x + 9600 = 0
Equation 2: y² - 190y + 9025 = 0
Equation 1: x² - 180x + 8100 = 0
Equation 2: y² - 170y + 7225 = 0
I. (y – 5)2 – 9 = 0
II. x2 – 3x + 2 = 0
I. 2y2 - 15y + 18 = 0
II. 2x2 + 9x - 18 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 41x² - 191x + 150 = 0
Equation 2: 43y² - 191y + ...
Equation 1: x² - 90x + 2025 = 0
Equation 2: y² - 88y + 1936 = 0
I. 3x6- 19x3+16=0
II. 9y4- 27y2+20=0
I. x2 – 39x + 360 = 0
II. y2 – 36y + 315 = 0
I. x2 + 16x + 63 = 0
II. y2 + 2y - 15 = 0
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