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I. 5x² - 24 x + 28 = 0 5x² - 14x - 10x + 28 = 0 x (5x – 14) – 2 (5x – 14 ) (x – 2 ) (5x – 14) x = 14/5, 2 II. 4y² - 8y - 12 = 0 4y²- 12y + 4y - 12 = 0 4 (y – 3) + 4(y – 3 ) (y– 3) (4y + 4) y = 3, -1 Hence, relationship between x and y cannot be established. 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 = 14/5, 2 If signs of quadratic equation is -ve and -ve respectively then the roots of equation will be +ve and -ve. (note: -ve sign will come in smaller root) So, roots of second equation = y = 3, -1 After comparing roots of quadratice eqution we can conclude that relation cannot be established between x and y.
l). p² - 29p + 204 = 0
ll). q² + 4q - 221 = 0
I. 20y² - 13y + 2 = 0
II. 6x² - 25x + 14 = 0
I. 27x² + 120x + 77 = 0
II. 56y² + 117y + 36 = 0
I. 22x² - 97x + 105 = 0
II. 35y² - 61y + 24 = 0
I). p2 = 81
II). q2 - 9q + 14 = 0
If x2 - 3x - 18 = 0 and y2 + 9y + 18 = 0, which of the following is true?
I. x2 + x – 42 = 0
II. y2 + 6y – 27 = 0
What is the nature of the roots of the quadratic equation x² – 5x + 7 = 0 ?
Solve the quadratic equations and determine the relation between x and y:
Equation 1: x² - 24x + 143 = 0
Equation 2: y² - 26y + 165 = 0