# Some weird equations

In our theoreticall class professor stated that from this equation $(C = constant)$

$$x^2 + 4Cx - 2Cy = 0$$

we can first get:

$$x = \frac{-4C + \sqrt{16 C^2 - 4(-2Cy)}}{2}$$

and than this one:

$$x = 2C \left[\sqrt{1 + \frac{y}{2C}} -1\right]$$

How is this even possible?

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This comes from the quadratic formula. See here for more details.

In your case, $$x^2 + 4Cx -2Cy = x^2 + 4Cx +(2C)^2 - (2C)^2-2Cy = (x+2C)^2 - (2C)^2 \left(1 + \dfrac{y}{2C} \right) = 0$$ Hence, we have that $$(x+2C)^2 = (2C)^2 \left(1 + \dfrac{y}{2C} \right) \implies x + 2C = \pm 2C \sqrt{\left(1 + \dfrac{y}{2C} \right)}$$ Hence, $$x = -2C \pm 2C \sqrt{\left(1 + \dfrac{y}{2C} \right)} = 2C \left( \pm \sqrt{\left(1 + \dfrac{y}{2C} \right)} - 1\right)$$ If you have a constraint that $x$ has the same sign as $C$ (for instance if $x,C > 0$), then $$x = 2C \left(\sqrt{\left(1 + \dfrac{y}{2C} \right)} - 1\right)$$

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Thank you. This was anwsered soooo fast i even had to wait to apply the anwser! Allso the topic you provided is very good. – 71GA Jan 14 '13 at 20:20

Factor out the $16 C^2$ from the square root and simplify.

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The secon one is just applying the general formula for second grade equations for x. From the second to the third you have to get out of the square root a common factor of $4C^2$, that becomes $2C$ when coming out of the root

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Here's the algebra: $$x^2 + 4Cx - 2Cy = (x+2C)^2-4C^2 - 2Cy = 0$$ Thus: $$(x+2C)^2 = 4C^2 + 2Cy = 2C(2C+y).$$ Take square roots: $$x_1 = -2C + \sqrt{2C(2C+y)} =\frac{-4C + \sqrt{16C^2 +8Cy}}{2}$$ and $$x_2 = -2C - \sqrt{2C(2C+y)}$$

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