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The correct answer is D
I. x2 – 7x + 12 = 0
II. y2 – 7y + 10 = 0
I. 6p² + 17p + 12 = 0
II. 12q² - 25q + 7 = 0
One of the roots of the equation p2 - (y+3)p + 6y = 0 is cube of 2. What will be the difference of other root and 'y'.
I.√(3x-17)+ x=15
II. y + 135/y=24
Solve the quadratic equations and determine the relation between x and y:
Equation 1: x² - 30x + 221 = 0
Equation 2: y² - 28y + 189 = 0
I. x2 + (9x/2) + (7/2) = - (3/2)
II. y2 + 16y + 63 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: x² - 32x + 207 = 0
Equation 2: y² - 51y + 648 = 0
I. x= √(20+ √(20+ √(20+ √(20…………….∞)) ) )
II. y= √(5√(5√(5√(5……….∞)) ) )
...I. 7p + 8q = 80
II. 9p – 5q = 57
Let 's' represent the sum of the highest root of equations I and III, and 'r' denote the product of the lowest root of equation I and the highest root o...