x² + 3x - 6y + y² = 9

Understand the Problem

The question presents an equation involving variables x and y: x² + 3x - 6y + y² = 9. It appears to be asking to analyze or manipulate this equation, possibly to rewrite it in a standard form (like that of a circle or other conic section), solve for one variable in terms of the other, or identify its graphical representation.

Answer

$(x + \frac{3}{2})^2 + (y - 3)^2 = \frac{81}{4}$

Steps to Solve

Rearrange the equation

Group the $x$ and $y$ terms together:

$$x^2 + 3x + y^2 - 6y = 9$$

Complete the square for x terms

To complete the square for $x^2 + 3x$, we need to add and subtract $(\frac{3}{2})^2 = \frac{9}{4}$:

$$x^2 + 3x + \frac{9}{4} - \frac{9}{4}$$ This gives us $(x + \frac{3}{2})^2 - \frac{9}{4}$.

Complete the square for y terms

To complete the square for $y^2 - 6y$, we need to add and subtract $(\frac{-6}{2})^2 = 9$:

$$y^2 - 6y + 9 - 9$$

This gives us $(y - 3)^2 - 9$.

Substitute back into the equation

Substitute the completed squares back into the original equation:

$$(x + \frac{3}{2})^2 - \frac{9}{4} + (y - 3)^2 - 9 = 9$$

Simplify the equation

Move the constants to the right side of the equation:

$$(x + \frac{3}{2})^2 + (y - 3)^2 = 9 + 9 + \frac{9}{4}$$

$$(x + \frac{3}{2})^2 + (y - 3)^2 = 18 + \frac{9}{4}$$

$$(x + \frac{3}{2})^2 + (y - 3)^2 = \frac{72}{4} + \frac{9}{4}$$

$$(x + \frac{3}{2})^2 + (y - 3)^2 = \frac{81}{4}$$

The equation is now in the standard form of a circle.

The equation in standard form is: $$(x + \frac{3}{2})^2 + (y - 3)^2 = \frac{81}{4}$$

More Information

The equation represents a circle with center $(-\frac{3}{2}, 3)$ and radius $\sqrt{\frac{81}{4}} = \frac{9}{2} = 4.5$.

Tips

A common mistake is forgetting to add the same value to both sides of the equation when completing the square. Also, mistakes can happen when calculating the values needed to complete the square, particularly with fractions.

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