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I have 2 equations: $$x^2 + y^2 = 12 \tag{1}$$ $$x^2 + y^2 -6x -2 \times 3^{1/2}y = 0 \tag{2}$$

On substituting the value of (1) in (2) we get: $$12 - 6x -2 \sqrt[3]{y} = 0 \tag{3}.$$

If we express $x$ in terms of $y$ we get: $$x = 12 - 2 \sqrt[2]{y}$$

This satisfies Eqn (3) but not for (2).

How is this is a root for equation (3) but not for (2)?

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  • $\begingroup$ are the equations $$x^2+y^2=12$$ and $$x^2+y^2-6y-2\cdot3^{1/3}y=0$$? $\endgroup$ – Dr. Sonnhard Graubner Sep 26 '17 at 16:57
  • $\begingroup$ On substituting 1 in 2 you get Eqn 3. $\endgroup$ – user33699 Sep 26 '17 at 17:03
  • $\begingroup$ Help with mathsJax is appreciated. $\endgroup$ – user33699 Sep 26 '17 at 17:03
  • $\begingroup$ the second is $$x^2+y^2-6x-2\cdot 3^{1/3}y=0$$ $\endgroup$ – Dr. Sonnhard Graubner Sep 26 '17 at 17:05
  • $\begingroup$ Nope its square root of 3 not cube root of 3. Also -1 rep was not by me . $\endgroup$ – user33699 Sep 26 '17 at 17:22
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Well, we have:

$$ \begin{cases} x^2+\text{y}^2=12\\ \\ x^2+\text{y}^2-6x-2\cdot\sqrt[3]{3}\cdot\text{y}=0 \end{cases}\tag1 $$

Notice that the second equation is a quadratic equation in $x$:

$$x^2+\text{y}^2-6x-2\cdot\sqrt[3]{3}\cdot\text{y}=x^2-6x-2\cdot\sqrt[3]{3}\cdot\text{y}+\text{y}^2=0\space\Longleftrightarrow\space$$ $$x=\frac{-\left(-6\right)\pm\sqrt{\left(-6\right)^2-4\cdot1\cdot\left(-2\cdot\sqrt[3]{3}\cdot\text{y}+\text{y}^2\right)}}{2\cdot1}=3\pm\sqrt{9+\text{y}\cdot\left(2\cdot\sqrt[3]{3}-\text{y}\right)}\tag2$$

So, we get:

$$x^2+\text{y}^2=\left(3\pm\sqrt{9+\text{y}\cdot\left(2\cdot\sqrt[3]{3}-\text{y}\right)}\right)^2+\text{y}^2=12\space\Longleftrightarrow\space$$ $$\text{y}\approx-1.88511\space\space\space\vee\space\space\space\text{y}\approx3.4471\tag3$$

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  • $\begingroup$ Its Square root if 3 not cube root of 3 . Also -1 rep wasnt from me. $\endgroup$ – user33699 Sep 26 '17 at 17:23

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