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P
Q16
My solutions to 16(c), which probably make a whole lot more sense because its JUST quadratic formula... You eventually derive y^2 + (1 - 2c)y + (c^2 - r^2) = 0 after substituting x = y^2 Now, y = (2c - 1 +-sqrt(1 - 4c + 4c^2 - 4c^2 + 4r^2 ))/2 -> y = (2c - 1 +-sqrt(1 - 4c + 4r^2))/2 Knowing...- Peter Zhang
- Post #140
- Forum: Mathematics
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