Question
(I) `p^2 + p = 56` (II) `q^2 - 17q
+ 72 = 0` In the following questions two equations numbered I and II are given. You have to solve both the equations. Give answer if; (1) p>q, (2) p<q, (3) p=q, (4) p ≥ q (5) p ≤ qSolution
I. `p^2 + p = 56` or `p^2 + p - 56 = 0` `p^2 + 8p - 7p - 56 = 0` or , (p – 7)(p + 8) = 0 Or p = - 8 or 7 , II. `q^2 - 17q + 72 = 0` , or (q – 8)(q – 9) = 0, or q = 8 or 9, Hence , q > p
Solve the quadratic equations and determine the relation between x and y:
Equation 1: x² - 5x + 6 = 0
Equation 2: y² - 7y + 12 = 0
I. Â 3y2Â + 13y - 16 = 0
II. 3x2 – 13x + 14 = 0
If x and y satisfy x² + y² = 25 and x + y = 7, find xy.
Roots of the quadratic equation 2x2 + x – 528 = 0 is
I. 3p² - 17p + 22 = 0
II. 5q² - 21q + 22 = 0
The quadratic equation (p + 1)x 2 - 8(p + 1)x + 8(p + 16) = 0 (where p ≠-1) has equal roots. find the value of p.
I. 64x2 - 64x + 15 Â = 0 Â Â Â Â
II. 21y2 - 13y + 2Â =0
I. 3y2 + 16y + 16 = 0
II. 2x2 + 19x + 45 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 21x² - 82x + 80 = 0
Equation 2: 23y² - 132y + 85 = 0
I. x2 + 11x + 30 = 0
II. y2 + 17y + 72 = 0
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