Quadratic EquationHard
Question
Let $M = 3x^{2} - 8xy + 9y^{2} - 4x + 6y + 13$, where $x,y \in R$, then
Options
A.M can not be equal to zero
B.M must be negative
C.$M > 2$
D.M must be positive
Solution
$M = 3x^{2} - 8yx + 9y^{2} - 4x + 6y + 15$
$$\begin{matrix} & \ = 2\left( x^{2} - 4xy + 4y^{2} \right) + \left( y^{2} + 6y + 9 \right) + \left( x^{2} - 4x + 4 \right) + 2 \\ & \ = 2(x - 2y)^{2} + (y + 3)^{2} + (x - 2)^{2} + 2 \\ & \ \Rightarrow \ M > 2 \end{matrix}$$
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