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#1 2008-10-21 03:47:02
- Danbee
- Member
- Registered: 2008-08-28
- Posts: 21
Solving a Quadratic-Quadratic System by Substitution
(x+5)² +(y+2)²=169
xy-21=0
Please help solve
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#2 2008-10-25 11:35:18
- justlookingforthemoment
- Moderator
- Registered: 2005-05-26
- Posts: 2,161
Re: Solving a Quadratic-Quadratic System by Substitution
If you think of this problem visually, you are trying to find the points on a graph where a hyperbola and a circle intersect.
1 -- (x+5)² +(y+2)²=169
2 -- xy - 21=0
xy - 21 = 0
xy = 21
y = 21/x
substitute y = 21/x into equation 1:
(x+5)² +(21/x+2)²=169
and expand:
x² + 10x + 25 + 441/(x²) + 84/x + 4 = 169
which gives:
x^4 + 10x³ - 140x² + 84x + 441 = 0
(x - 7)(x³ + 17x² - 21x - 63) = 0
solve for x:
∴ x ≈ -17.9734 or x ≈ -1.44775 or x ≈ 2.42112 or x = 7
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