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Start learning 50% faster. Sign in nowFrom I: x2 – 39x + 360 = 0 Or, x2 – 15x – 24x + 360 = 0 Or, x × (x – 15) – 24 × (x – 15) = 0 Or, (x – 15) × (x – 24) = 0 So, x = 15 or, x = 24 From II: y2 – 36y + 315 = 0 Or, y2 – 15y – 21y + 315 = 0 Or, y × (y – 15) – 21 × (y – 15) = 0 Or, (y – 15) × (y – 21) = 0 So, y = 15 or y = 21 So, the relationship cannot be established. Hence, option c.
I. x² - 33x + 270 = 0
II. y² - 41y + 414 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 41x² - 191x + 150 = 0
Equation 2: 43y² - 191y +...
I. 88x² - 13 x – 56 = 0
II. 15 y² + 41 y + 28 = 0
I. 2y2 + 31y + 99 = 0
II. 4x2 + 8x – 45 = 0
I. 2p2 - 3p – 2 = 0 II. 2q2 - 11q + 15 = 0
I. x2 + (9x/2) + (7/2) = - (3/2)
II. y2 + 16y + 63 = 0
I). 4p + 3q = 25
II). 5p + 2q = 26
I. 6x2 – 7x - 20 = 0
II. 3y2 - y - 14 = 0
What is the nature of the roots of the quadratic equation x² – 5x + 7 = 0 ?
I. 15b2+ 26b + 8 = 0
II. 20a2+ 7a - 6 = 0