1988 Virginia Slims of Houston – Doubles
Kathy Jordan and Martina Navratilova were the defending champions but only Navratilova competed that year with Lori McNeil.
Doubles | |
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1988 Virginia Slims of Houston | |
1987 Champions | ![]() ![]() |
Champions | ![]() ![]() |
Runners-up | ![]() ![]() |
Final score | 6–7, 6–2, 6–4 |
McNeil and Navratilova lost in the final 6–7, 6–2, 6–4 against Katrina Adams and Zina Garrison.
Seeds
Champion seeds are indicated in bold text while text in italics indicates the round in which those seeds were eliminated.
Lori McNeil / Martina Navratilova (Final) Katrina Adams / Zina Garrison (Champions) Rosalyn Fairbank / Candy Reynolds (Quarterfinals) Penny Barg / Mercedes Paz (Semifinals)
Draw
Key
- Q = Qualifier
- WC = Wild Card
- LL = Lucky Loser
- Alt = Alternate
- SE = Special Exempt
- PR = Protected Ranking
- ITF = ITF entry
- JE = Junior Exempt
- w/o = Walkover
- r = Retired
- d = Defaulted
First Round | Quarterfinals | Semifinals | Final | ||||||||||||||||||||||||
1 | ![]() ![]() | 2 | 6 | 6 | |||||||||||||||||||||||
![]() ![]() | 6 | 2 | 4 | 1 | ![]() ![]() | 6 | 6 | ||||||||||||||||||||
![]() ![]() | 3 | 5 | ![]() ![]() | 1 | 3 | ||||||||||||||||||||||
![]() ![]() | 6 | 7 | 1 | ![]() ![]() | 7 | 6 | |||||||||||||||||||||
3 | ![]() ![]() | 6 | 6 | ![]() ![]() | 6 | 2 | |||||||||||||||||||||
![]() ![]() | 1 | 3 | 3 | ![]() ![]() | 2 | 2 | |||||||||||||||||||||
![]() ![]() | 6 | 6 | ![]() ![]() | 6 | 6 | ||||||||||||||||||||||
![]() ![]() | 1 | 4 | 1 | ![]() ![]() | 7 | 2 | 4 | ||||||||||||||||||||
![]() ![]() | 4 | 4 | 2 | ![]() ![]() | 6 | 6 | 6 | ||||||||||||||||||||
![]() ![]() | 6 | 6 | ![]() ![]() | 3 | 4 | ||||||||||||||||||||||
![]() ![]() | 3 | 1 | 4 | ![]() ![]() | 6 | 6 | |||||||||||||||||||||
4 | ![]() ![]() | 6 | 6 | 4 | ![]() ![]() | 1 | 4 | ||||||||||||||||||||
![]() ![]() | 6 | 7 | 4 | 2 | ![]() ![]() | 6 | 6 | ||||||||||||||||||||
![]() ![]() | 7 | 6 | 6 | ![]() ![]() | 1 | 1 | |||||||||||||||||||||
![]() ![]() | 6 | 1 | 2 | ![]() ![]() | 6 | 6 | |||||||||||||||||||||
2 | ![]() ![]() | 7 | 6 |
gollark: So you can do `2x^4+ px^3 - 6x^2 + qx + 6 = ax^4 + (b-a)x^3 + (c-b)x^2 + (d-c)x - d`, and you know the coefficients on x^4 and so on should be equal.
gollark: Which you can then simplify to ax^4 + (b-a)x^3 + (c-b)x^2 + (d-c)x - d.
gollark: ax^4 + bx^3 + cx^2 + dx - ax^3 - bx^2 - cx - d
gollark: If you expand/simplify (x-1)(ax^3+bx^2+cx+d) you get, er, a lot of things, hold on.
gollark: This is a way you can do that, though.
References
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