Question
In each of these questions, two equations (I) and (II)
are given.You have to solve both the equations and give answer I. 3x² –7x + 4 = 0 II. 2y² – 9y + 10 = 0Solution
I. 3x² –7x + 4 = 0 ⇒ 3x²– 4x - 3x +4 = 0 ⇒ (3x – 4) (x -1) = 0 x = 4/3 or 1 II. 2y² - 9y + 10 = 0 ⇒ 2y² - 4y - 5y + 10 = 0 ⇒ (2y - 5) (y -2) =0 ⇒ y = 5/2 or 2 Hence, y > x Alternate Method: if signs of quadratic equation is -ve and +ve respectively then the roots of equation will be +ve and +ve. So, roots of first equation = x = 4/3, 1 So, roots of second equation = y = 5/2, 2 After comparing roots of quadratice eqution we can conclude that y > x.
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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