Question
For this equation, x is real, b and c are each real constants, and b is less than c. Find the possible values of y so that y = (x^2-bc)/(2x-b-c)
Answers
oobleck
the only exclusion is that 2x ≠ b+c
Other than that, you can pick any x,b,c you choose.
What else are you looking for?
Maybe you can play around with this.
If m = (c+b)/2 and n = (c-b)/2 then
b = m - n
c = m + n
and y = (x^2 - (m^2-n^2))/((x-b)+(x-c))
Other than that, you can pick any x,b,c you choose.
What else are you looking for?
Maybe you can play around with this.
If m = (c+b)/2 and n = (c-b)/2 then
b = m - n
c = m + n
and y = (x^2 - (m^2-n^2))/((x-b)+(x-c))