Too many (or not any) values?

Relevant wiki: Completing the Square - Applications

Let's separate each of the terms into x's, y's, and constants: $x^2-2x+y^2-3y+10=0$ Then, complete the square: $(x^2-2x+1)-1+(y^2-3y+2.25)-2.25+10=0$ $\implies (x-1)^2+(y-1.5)^2+6.75=0$ $\implies (x-1)^2+(y+1.5)^2=-6.75$ Now, look at the equation above. It states 2 positive numbers (because they are squared) add up to a negative number. However, assuming $x$ and $y$ are real numbers, this is impossible. Therefore, the answer is 0. $\beta_{\lceil \mid \rceil}$

I offer two solutions:

1.

Consider the equation as a quadratic in terms of $y$ : $y^2-3y+(x^2-2x+10) = 0$ Consider the discriminant of this quadratic: $9-4(x^2-2x+10) = -4x^2 + 8x - 31)$ Now, considering the quadratic in $x$ above, the discriminant is $64-4(-4)(-31) = 64 - 496 = -432$ Since the discriminant is negative, there are no real roots. And since the leading coefficient is negative, the value of the quadratic must be negative for all real $x$ . Thus, for all real $x$ , the discriminant of our original quadratic (in $y$ ) is also negative. Finally, this implies that there are no real $y$ satisfying our original equation, so the answer must be $\boxed{0}$ .

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2.

Consider the equation as that of an circle. We complete the square for both $x$ and $y$ : $\begin{aligned} (x^2 - 2x + 1) - 1 + (y^2 - 3y + \frac{9}{4}) - \frac{9}{4} +10 &= 0 \\ (x-1)^2 + (y-2)^2 &= -\frac{27}{4} \end{aligned}$

This is the standard form of an circle: $(x-x_0)^2 + (y-y_0)^2 = r^2$ except, in our case, $r^2 = -\frac{27}{4}$ .This is a circle with an imaginary radius(!) and therefore no real $x$ or $y$ can satisfy this equation.