Question
I. 3x2 – 16x + 21 = 0 II.
y2 – 13y + 42 = 0 In each question two equations numbered (I) and (II) are given. You should solve both the equations and mark appropriate answer.Solution
I. 3x2 – 16x + 21 = 0 => 3x2 – 9x - 7x + 21 = 0 => 3x (x – 3) – 7 (x – 3) = 0 => x = 3, 7/3 II. y2 – 13y + 42 = 0 => y2 – 6y – 7y + 42 = 0 => y (y – 6) – 7(y – 6) = 0 => y = 7, 6 Hence, x < y 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 = 7/3, 3 So, roots of second equation = y = 7, 6 After comparing we can conclude that x < y.
I. x2 - 5x - 14 = 0
II. y2 - 16y + 64 = 0
I. 22x² - 97x + 105 = 0
II. 35y² - 61y + 24 = 0
- If the quadratic equation x² + 18x + n = 0 has real and equal roots, what is the value of n?
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 27x² - 114x + 99 = 0
Equation 2: 18y² - 70y + 68 = 0
I. 56x² - 99x + 40 = 0
II. 8y² - 30y + 25 = 0
I). p2 - 26p + 165 = 0
II). q2 + 8q - 153 = 0
I. x2 + 91 = 20x
II. 10y2 - 29y + 21 = 0
In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y.
I. x
I. 8x – 3y = 85
II. 4x – 5y = 67
I. 6x2 - 41x+13=0
II. 2y2 - 19y+42=0
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