amaths quad equation questions

• Bigcable22

  12 Mar 08, 21:09

  

   

  Qns 10,11b-d,12 how to do arh??

• eagle

  12 Mar 08, 21:14

  10) Find the range of values of c for which x^2 + 6x - 5 is greater than 8x + c

  So

  x^2 + 6x - 5 > 8x + c
  x^2 - 2x - (5+c) > 0

  so b^2 - 4ac > 0
  4 + 4(1)(5+c) > 0
  1 + 5 + c > 0

  c > -6

• eagle

  12 Mar 08, 21:15

  The rest are similar ways of doing... playing with the determinant... you try first...

• Bigcable22

  12 Mar 08, 21:23

  Originally posted by eagle:

  10) Find the range of values of c for which x^2 + 6x - 5 is greater than 8x + c

  So

  x^2 + 6x - 5 > 8x + c
  x^2 - 2x - (5+c) > 0

  so b^2 - 4ac > 0
  4 + 4(1)(5+c) > 0
  1 + 5 + c > 0

  c > -6

  ok i got it but cannot do the rest except 11 a.) .......................................................................................................................

• ^tamago^

  12 Mar 08, 21:24

  10. x²+6x-5 > 8x+c
  x²-2x-5 > c
  (x²-2x+1)-6 > c
  (x-1)²-6 > c
  (x-1)²-6 is minimum when x=1, i.e. c < -6
  ∴ Solution set is given by { c: c ∈ R, c < -6 }
  
  10. x²+6x-5 > 8x+c (alternative, better solution)
  x²-2x-(5+c) > 0
  For the line to not meet the curve, the solution must not have any real root(s), i.e. b²-4ac < 0
  (-2)²+4(1)(5+c) < 0
  4c < -24
  c < -6
  ∴ Solution set is given by { c: c ∈ R, c < -6 }
  
  
  
  11b) x+3y=k-1 ⎯⎯ (1)
  y²=2x+5 ⎯⎯ (2)
  sub (2) into (1): ½(y²-5)+3y=k-1
  y²+6y-(3+2k)=0
  For the line to meet the curve, the solution must have real root(s), i.e. b²-4ac ≥ 0
  6²+4(1)(3+2k) ≥ 0
  48+8k ≥ 0
  k ≥ -6
  ∴ Solution set is given by { k: k ∈ R, k ≥ -6 }
  
  11c) y² = 8x-x² ⎯⎯ (3)
  y = kx+2 ⎯⎯ (4)
  sub (3) into (4): (kx+2)² = 8x-x²
  k²x²+4kx+4=8x-x²
  (k²+1)x²+(4k-8)x+4=0
  For the line to meet the curve at 2 points, the solution must have 2 real root(s), i.e. b²-4ac > 0
  (4k-8)²-16(k²+1) > 0
  64k < 48
  k < ¾
  ∴ Solution set is given by { k: k ∈ R, k < ¾ }
  
  You can read and try to do 11d) and 12. (:
  
  
  
  
• Bigcable22

  13 Mar 08, 00:52

  Originally posted by ^tamago^:

  10. x²+6x-5 > 8x+c
  x²-2x-5 > c
  (x²-2x+1)-6 > c
  (x-1)²-6 > c
  (x-1)²-6 is minimum when x=1, i.e. c < -6
  Solution set is given by { c: c ∈ R, c < -6 }
  
  
  11b) x+3y=k-1 ⎯⎯ (1)
  y²=2x+5 ⎯⎯ (2)
  sub (2) into (1): ½(y²-5)+3y=k-1
  y²+6y-(3+2k)=0
  For the line to meet the curve, the solution must have real root(s), i.e. b²-4ac ≥ 0
  6²+4(1)(3+2k) ≥ 0
  48+8k ≥ 0
  k ≥ -6
  Solution set is given by { k: k ∈ R, k ≥ -6 }
  
  11c) y² = 8x-x² ⎯⎯ (3)
  y = kx+2 ⎯⎯ (4)
  sub (3) into (4): (kx+2)² = 8x-x²
  k²x²+4kx+4=8x-x²
  (k²+1)x²+(4k-8)x+4=0
  For the line to meet the curve at 2 points, the solution must have 2 real root(s), i.e. b²-4ac > 0
  (4k-8)²-16(k²+1) > 0
  64k < 48
  k < ¾
  Solution set is given by { k: k ∈ R, k < ¾ }
  
  You can read and try to do 11d) and 12. (:
  
  
  

  i dont quite understand your 11.b)..  especially how you eliminate the  y^2=2x+5
    and sub the answer into (1)

• ^tamago^

  13 Mar 08, 00:56

  Originally posted by Bigcable22:

  i dont quite understand your 11.b)..  especially how you eliminate the  y^2=2x+5
    and sub the answer into (1)

  
  1b) x+3y=k-1 ⎯⎯ (1)
  y²=2x+5 ⎯⎯ (2)
  2x=y²-5 ⎯⎯ (3)
  sub (3) into (1): ½(y²-5)+3y=k-1