Quadratic Equation Question Intensity
May 9th, 2008

Question:

I have figured out the equation for a demand curve by plotting some points, and I found it to be: y = 2E-05x^2 - 0.195x + 716.67 Expressed in the usual way, this is: y = 0.00002x^2 - 0.195x + 716.67

Now, the problem is this... I don't know how to solve this equation for x? This is as far as I get:

y - 716.67 = 0.00002x^2 - 0.195x
y - 716.67 = x(0.00002x - 0.195)

Then I'm stuck. How do I get the other x out of the bracket??? The aim is to have an equation with x on one side and all the other stuff on the other side.

Answer:

y = 0.00002x^2 - 0.195x + 716.67

This is not a quadratic equation in the common sense of the word, as you have an x and a y. therefore, if you want x = to something, it will include a y. What you may want to find though is where it crosses the axis:

when x = 0 ,
y=716.67
BUT:
The real equation for quadratic equations which can be found at http://www.grabthebasics.com/quad/quad7.asp is
(If you want to know why this is true, then see http://www.geocities.com/dirkie6/page4.html)
When:
y = 0.00002x^2 - 0.195x + 716.67
b^2-4ac=0.195^2-4*716.67*0.00002
= -0.535311
If we put this into the above equation, this means the we are square-rooting a negative number
This means, that the equation that you gave me,does not have real roots when y=0. It has imaginary roots.The solutions which are imaginary can be found at http://members.tripod.com/~kselva/quad.html
If I am totally misunderstanding your question, and y is a constant, then according to your equation,
y = 0.00002x^2 - 0.195x + 716.67
and therefore
0.00002x^2 - 0.195x + 716.67-y=0
which could have routes if b^-4a(c-y)>0
BUT THERE WAS MORE>>>>>

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