Square of sum: Using variables

To determine the correct relationship between $x$ and $y$, let's transform each equation:

Step 1: Transform the First Equation

The first equation is $x^2 + 4 = -6y$. Rearranging gives us:

$x^2 = -6y - 4$

Now, aim to complete the square for expressions involving $x$ and $y$.

Step 2: Transform the Second Equation

The second equation is $y^2 + 9 = -4x$. Rearranging gives us:

$y^2 = -4x - 9$

Step 3: Complete the Square

Let's complete the square for the terms $x^2$ and $y^2$.

For $x^2 = -6y - 4$:

$x^2 + 4x + 4 = -6y$

Thus, it becomes:

$(x + 2)^2 = -6y$

For $y^2 = -4x - 9$:

$y^2 + 6y + 9 = -4x$

Thus, it becomes:

$(y + 3)^2 = -4x$

Step 4: Combine and Analyze

Substitute back to express a sum of squares:

Adding these completes the square:

$(x + 2)^2 + (y + 3)^2 = 0$

This result shows that both squares, squared terms are zero-sum, revealing the conditions under which equations balance.

Thus, the correct choice according to the transformations conducted is:

$(x+2)^2 + (y+3)^2 = 0$

Therefore, the solution to the problem is $(x+2)^2 + (y+3)^2 = 0$.

To solve this problem, we will follow these steps:

• Step 1: Identify the equations and express one variable in terms of the other.
• Step 2: Substitute into the other equation and simplify.
• Step 3: Perform calculations to solve for the variable.
• Step 4: Use the solution to find the second variable.

Let's work through the solution together:

Step 1: Given $xy = 9$, express $y$ as $\frac{9}{x}$.

Step 2: Substitute into the first equation:

$\sqrt{x} + \sqrt{\frac{9}{x}} = \sqrt{\sqrt{61} + 6}$.

Step 3: Simplify this equation. Let $a = \sqrt{x}$ and $b = \sqrt{y}$.

Then, $a + b = \sqrt{\sqrt{61} + 6}$ and $ab = \sqrt{9} = 3$.

Squaring both sides of the linear equation:

$(a + b)^2 = \sqrt{61} + 6$.

$a^2 + 2ab + b^2 = \sqrt{61} + 6$.

Using $ab = 3$, we get $2ab = 6$.

This leads to $a^2 + b^2 = \sqrt{61}$.

Replacing $a = \sqrt{x}$ and $b = \sqrt{y}$:

Let $a^2 = x$ and $b^2 = y$ and use the identity $(a - b)^2 = a^2 + b^2 - 2ab = \sqrt{61} - 6$.

So, $a - b = \sqrt{\sqrt{61} - 6}$.

Now let $S = a + b$ and $P = ab$ from previous steps.

From $S = \sqrt{\sqrt{61} + 6}$ and $P = 3$, solve: $t^2 - St + P = 0$.

This quadratic in $t$ gives solutions $t = \frac{S \pm \sqrt{S^2 - 4P}}{2}$.

The quadratic roots are $a = \frac{\sqrt{61} + 6}{2} \pm \frac{5}{2}$ and $b = \frac{\sqrt{61} + 6}{2} \mp \frac{5}{2}$.

Thus, $x = a^2 = (\frac{\sqrt{61}}{2} + 2.5)^2$ or $(\frac{\sqrt{61}}{2} - 2.5)^2$.

Similarly for $y$.

Therefore, the solutions are:

$x = \frac{\sqrt{61}}{2} - 2.5$, $y = \frac{\sqrt{61}}{2} + 2.5$

or

$x = \frac{\sqrt{61}}{2} + 2.5$, $y = \frac{\sqrt{61}}{2} - 2.5$.

We know that:

$\frac{AB}{BC}=\sqrt{\frac{x}{2}}$

We also know that AB equals X.

First, we will substitute the given data into the formula accordingly:

$\frac{x}{BC}=\frac{\sqrt{x}}{\sqrt{2}}$

$x\sqrt{2}=BC\sqrt{x}$

$\frac{x\sqrt{2}}{\sqrt{x}}=BC$

$\frac{\sqrt{x}\times\sqrt{x}\times\sqrt{2}}{\sqrt{x}}=BC$

$\sqrt{x}\times\sqrt{2}=BC$

Now let's look at triangle ABC and use the Pythagorean theorem:

$AB^2+BC^2=AC^2$

We substitute in our known values:

$x^2+(\sqrt{x}\times\sqrt{2})^2=m^2$

$x^2+x\times2=m^2$

Finally, we will add 1 to both sides:

$x^2+2x+1=m^2+1$

$(x+1)^2=m^2+1$