Question 329607
I believe you are saying:


(1/4)*x^2 + (17/4) = 2*x


If this is not the case, then send me an email with a clarification.


You can start by subtracting 2*x from both sides of the equation to get:


(1/4)*x^2 - 2*x + (17/4) = 0


Multiply both sides of this equation by 4 to get:


x^2 - 8*x + 17 = 0


Complete the squares on the (x^2 - 8*x) part to get:


(x-4)^2 - 16 + 17 = 0


Add 16 and subtract 17 from both sides of this equation to get:


(x-4)^2 = -1


Take the square root of both sides of this equation to get x-4 = +/- square root of (-1).


Since square root of (-1) is not a real value, then this quadratic equation does not have any real roots.


It has imaginary, or complex roots, complex being a combination of real and imaginary.


A graph of your original equation is shown below:


{{{graph(600,600,-10,10,-10,10,(1/4)*x^2 - 2*x + (17/4))}}}


You can see that the graph of this equation does not cross the x-axis, therefore this equation doesn't have any real roots.


The roots that we did find are:


x = 4 +/- square root of (-1) 


Since square root of (-1) is equal to i, then the roots become:


x = 4 +/- i.


Once again, these are not real roots.


These are complex roots, because they are a combination of a real part (the 4) and an imaginary part (the i, where i = square root of (-1)).


If you had used the quadratic formula,  you would have gotten the same answer.


The quadratic formula is x = (-b +/- square root of (b^2-4ac) / (2a).


In your equation of x^2 - 8*x + 17 = 0,


a = 1
b = -8
c = 17


The quadratic formula becomes:


x = (-(-8) +/- square root of (64-4*1*17)) / (2*1) which becomes:


x = (8 +/- square root of (64 - 68)) / 2 which becomes:


x = (8 +/- square root of (-4) / 2 which becomes:


x = (8 +/- square root of (4*-1) / 2 which becomes:


x = (8 +/- 2 * square root of (-1) / 2 which becomes:


x = 4 +/- square root of (-1).


Since square root of (-1) is equal to i, then the solution becomes


x = 4 +/- i.


This is the same answer we derived using the completing the squares method.