Quadratic Equations

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Directions (Q. 1-5): in each question two equations numbered I and II are given. You have to solve both the equations and mark the answer.

  1. a)If x > y
  2. b)If x ≥ y
  3. c)If x < y
  4. d)If x ≤ y
  5. e)If x = y or no relation can be established between x and y.

1).

  1. x2– 16 = 0
  2. y4–  {[(8×2)9/4] / √y } = 0

2).

  1. x2– 2√3x – 72 = 0
  2. y2+ 9√3x + 60 = 0

3).

  1. x6– [(√(54×6))6.5/ √x] = 0
  2. y – (36 × 81)1/2= 0

4).

  1. x2– 17x – 234 = 0
  2. 4y2+ 32y – 192 = 0

5).

  1. x2= 784
  2. y2– 11y – 102 = 0

 

Explanation With Answer Key:

1). D)

  1. x2– 16 = 0

x2 = 16

x = ±4

  1. y4–  {[(8×2)9/4] / √y } = 0

y√y  –  [(42)9/4]  = 0

(4+1/2) – (4)9/2 = 0

y9/2 – 49/2 = 0

y = 4

Hence x ≤ y

2). B)

  1. x2– 2√3x – 72 = 0

x– 6√3x + 4√3x – 72 = 0

x (x – 6√3) + 4√3 (x – 6√3) = 0

x = -4√3,  6√3

  1. y2+ 9√3y + 60 = 0

y2 + 4√3y + 5√3y + 60 = 0

y(y + 4√3) + 5√3 (y + 4√3) = 0

y = -4√3,  -5√3

Hence x ≥ y

3). C)

  1. x6– [(√(54×6))6.5/ √x] = 0

x6+(1/2) – (18)6.5 = 0

x6.5 = 186.5

x = 18

  1. y – (36 × 81)1/2= 0

y – (6 × 9) = 0

y – 54 = 0

y = 54

Hence x < y

4). E)

  1. x2– 17x – 234 = 0

x2 – 26x + 9x – 234 = 0

x(x – 26) + 9 (x – 26) = 0

(x + 9) (x – 26) = 0

x = -9, 26

  1. 4y2+ 32y – 192 = 0

y2 + 32y – 768 = 0

y2 + 48y – 16y – 768 = 0

y (y + 48) -16 (y + 48)

y = -48, 16

Hence relation cannot be determined

5). E)

  1. x2= 784

x = ±28

  1. y2– 11y – 102 = 0

y2 – 17y + 6y – 102 = 0

y (y – 17) +6 (y – 17) = 0

y = 17, -6

Hence Relation cannot be determined

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