Question 820654
In the case of x^2 - 2x + c = 0, it is of the form ax^2 + bx + c = 0 where a = 1, b = -2 and c is unknown.



The equation ax^2 + bx + c = 0 has at most 2 solutions. The type and number of unique solutions are determined by the equation below


D = b^2 - 4ac


which is the discriminant formula. The rules are this


* If D > 0, then you'll have 2 real solutions that are distinct.


* If D = 0, then you'll have exactly 1 real solution (that's also rational).


* If D < 0, then you'll have 2 imaginary solutions (aka, complex solutions)



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So from the rules above, we will have 2 real solutions when D > 0.


D > 0


b^2 - 4ac > 0  ... replace D with b^2 - 4ac (since D = b^2 - 4ac)


(-2)^2 - 4(1)c > 0 ... plug in a = 1, b = -2, and leave c alone


4 - 4c > 0


4 - 4c + 4c > 0 + 4c


4 > 4c


4c < 4


4c/4 < 4/4


c < 1


So if c < 1, then D > 0 which will lead to 2 distinct real solutions.


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Now let's move onto the second case: When D = 0



D = 0


b^2 - 4ac = 0


(-2)^2 - 4(1)c = 0


4 - 4c = 0


4 = 4c


4c = 4


c = 1


If c = 1, then you'll get exactly one real solution.


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You can probably guess what the last answer must be, but let's work through it


D < 0


b^2 - 4ac < 0


(-2)^2 - 4(1)c < 0


4 - 4c < 0


4 < 4c


4c > 4


c > 1


So if c > 1, then you'll get 2 imaginary solutions.