Question
I. √(74x-250 )– x=15 II. √(3y²-37y+18)+
2y=18 In the following questions, two equations numbered I and II are given. You have to solve both the equations and find out the correct option. Give answerÂSolution
I. √(74x-250 )– x=15  √(74x-250 )=15+x On squaring both sides, we get 74x-250=225+x²+30x x²-44x+475=0 x²-25x-19x+475=0 x (x-25)- 19 (x-25)= 0 (x-19)(x-25)= 0 x=19,25 II. √(3y²-37y+18)+ 2y=18 √(3y²-37y+18)=18-2y On squaring both the sides, we get 3y²-37y+18=324+4y²-72y y²-35y+306=0 y²-17y-18y+306=0 y(y-17)- 18(y-17)= 0 (y-17)(y-18)= 0 y=17,18 Hence, x>y
I. 3p² - 17p + 22 = 0
II. 5q² - 21q + 22 = 0
I. 3x2 – 17x + 10 = 0
II. y2 – 17y + 52 = 0
I. 35 y² + 58 y + 24 = 0
II. 21 x² + 37 x + 12 = 0
I. 5x2 – 7x – 6 =0
II. 2y2 – 5y – 7 =0
I. 40 x² - 93 x + 54 = 0
II. 30 y² - 61 y + 30 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 37x² - 172x + 135 = 0
Equation 2: 29y² - 132y + ...
I: x2Â + 31x + 228 = 0
II: y2 + 3y – 108 = 0
I. 14p2 – 135p + 81 = 0
II. 7q2 – 65q + 18 = 0
I. 64x2 - 64x + 15 Â = 0 Â Â Â Â
II. 21y2 - 13y + 2Â =0
I. 6y2 – 23y + 20 = 0
II. 4x2 – 24 x + 35 = 0
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