General Question
What is x, a and b of the following problem?
For which a and b are the solutions of the equation -4bx^2 + (a-2)x +a +b = 0 imaginary. (a,b are real numbers)
I have come to the point where I evaluated a to be 2, but am stuck with evaluating b.
Here’s the a process:
x² – ( (a-2) / 4b )x – ( (a+b) / 4b) = 0 //divided by -4b
x² – ( (a-2) / 4b )x + ((a-2)² / 64b²) = ((a+b)/4b) + ((a-2)² / 64b²) //completing the square
(x – (a-2)/8b)² = (16b(a+b) + (a-2)²) / 64b²
x = ±root(16b(a+b)+(a-2)²) / 8b + (a-2) / 8b =
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from this follows that a-2 need be 0 (or a = 2) because the real part of the complex solution needs be zero (remember, we’re looking for a solution only with imaginary part).
any suggestion as to how to proceed with solving for b?
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