Set-20 Quadratic Equations (Inequalities) For SBI PO and SBI Clerk 2019 | Must Go Through These Questions

• x> y
• x<y
• x ≤ y
• x ≥ y
• x =y Or cannot be determined

10.

• Directions:1-5) In the following questions, two equations numbered are given in variables a and b. you have to solve both the equations and find out the relationship between a and b. Then give answer accordingly-

  1. Question

  I.5a² – 18a + 9 =0
  II.3b² + 5b – 2 = 0

  • If a > b
  • If a < b
  • If a ≥ b
  • If a ≤ b
  • If a = b or relation cannot be established
  Ans: 1
  5a² – 18a + 9  = 0
  => 5a² – 15a – 3a  + 9 =0
  => (5a – 3 )(a-3 )= 0
  => a=  3/ 5 or  a= 3
  3b² + 5b – 2 = 0
  => 3b² + 6b – b -2 = 0
  => (3b-1)(b + 2) = 0
  => b = 1/3  or  -2

• 2. Question

  I.√a – √6 / √a = 0
  II. b³ – 6^3/2 = 0

  • If a > b
  • If a < b
  • If a ≥ b
  • If a ≤ b
  • If a = b or relation cannot be established
  Ans: 5
  √a –   √6 / √a = 0
  a – √6 = 0
  a = √6
  b³ – 6^(3/2) = 0
  =>b³ = (√6)³
  => b = √6There is no relation between a and b

• 3. Question

  .(625)^1/4a + √1225 = 155
  II.√196b + 13 = 279

  • If a > b
  • If a < b
  • If a ≥ b
  • If a ≤ b
  • If a = b or relation cannot be established
  Ans: 1
  5a + 35 = 155
  => 5a = 155 – 35
  => a = 120/ 5 = 24√196 b + 13 = 279
  => 14b = 279 -13
  => b = 266/14 =19

  From I and II, a>b

• 4. Question

  I. 3a² – 17a + 24 =0
  II. 4b² – 15b + 14 = 0

  • If a > b
  • If a < b
  • If a ≥ b
  • If a ≤ b
  • If a = b or relation cannot be established
  Ans: 1
  3a² -17a  + 24 =0
  => 3a² – 9a  – 8a +24 = 0
  => (3a- 8)(a-3) = 0
  =>  a=  8/3 or 3
  4b² – 15b + 14 = 0
  => 4b² – 8b -7b +14 = 0
  => (4b -7 )(b-2)=0
  => b = 7 /4 or  2From I and II, a>b

• 5. Question

  I. a² – 2a – √5 a + 2√5 = 0
  II. b² – √3 b – √2 b + √6 =0

  • If a > b
  • If a < b
  • If a ≥ b
  • If a ≤ b
  • If a = b or relation cannot be established
  Ans: 1
  a² – 2a- √5 a + 2√5 = 0
  => a(a-2 ) – √5 (a-2 )= 0
  => (a-2)(a-√5)=0
  => a= 2 or √5
  b² – √3 b  – √2 b + √6 =0
  => b(b-√3) – √2(b – √3) = 0
  => (b – √2)(b- √3) =0
  => b = √2 or √3From I and II, a>b

  Directions:6-10) In each of the following questions, two equations (i) and (ii) are given. You have to solve them and find the correct option.

  6. Question

  a)3+1/x-10/x²=0
  b)3y²-5y=12

  • x> y
  • x<y
  • x ≤ y
  • x ≥ y
  • x =y Or cannot be determined
  Ans: 5
  a)3+1/x-10/x²=03+1/x-10/x²=0

  3x²+x-10=0

  (3x-5)(x+2)=0

  x=5/3 and -2

  b)3y²-5y=12

  3y²-5y-12=0

  (3y+4)(y-3)=0

  y= 3 and -4/3  

  Hence, relationship cannot be determined

• 7. Question

  a)3y-4x+5=0
  b) 6x-5y=11

  • x> y
  • x<y
  • x ≤ y
  • x ≥ y
  • x =y Or cannot be determined
    Ans: 1
  3y-4x+5=0………………….. (1)6x-5y=11…………………… (2)

  Multiply equation (i) by3

  9y-12x+15=0

  And multiply equation (ii) by 2

  12x-10y=22

  Solving these equations, we get

  x=-4 and y=-7

  Hence x>y

• 8. Question

  a)7x²+21x+14=0
  b)6y²-35y-26=0

  • x> y
  • x<y
  • x ≤ y
  • x ≥ y
  • x =y Or cannot be determined
  Ans: 2
  a)7x²+21x+14=07x²+14x+7x+14=0

  (x+1)(7x+14)=0

  x=-1 and -2

  b)6y²-35y-26=0

  6y²+4y-39y-26=0

  2y(3y+2)-13(3y+2)=0

  (2y-13)(3y+2)=0

  y=13 /2 and -2/3

  y= 6.5 and -0.6

  Hence x<y

• 9. Question

  a)15x²-61x+56=0
  b)20y²+51y+28=0

  • x> y
  • x<y
  • x ≤ y
  • x ≥ y
  • x =y Or cannot be determined
  Ans: 1
  a)15x²-61x+56=015x²-40x-21x+56=0

  (5x-7)(3x-8)=0

  x=8/3 and 7/5

  b)20y²+51y+28=0

  20y²+16y+35y+28=0

  (4y+7)(5y+4)=0

  y=-4/5 and -7/4

  Hence, x>y

• 10. Question

  a)4x²-17x+18=0
  b)2y²+13y+21=0

  • x> y
  • x<y
  • x ≤ y
  • x ≥ y
  • x =y Or cannot be determined
  Ans: 1
  a)4x²-17x+18=04x²-8x-9x+18=0

  (4x-9)(x-2)=0

  x=9/4 and 2

  b)2y²+13y+21=0

  2y²+7y+6y+21=0

  (y+3)(2y+7)=0

  y=-3 and -7/2

  Hence x>y