Square of sum: Using variables

Given the rectangle ABCD

AB=X

The ratio between AB and BC is $\sqrt{\frac{x}{2}}$

We mark the length of the diagonal A the rectangle in m

Check the correct argument:

Given that:

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

Given that AB equals X

We will substitute accordingly in the formula:

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

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

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

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

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

Now let's focus on triangle ABC and use the Pythagorean theorem:

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

Let's substitute the known values:

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

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

We'll add 1 to both sides:

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

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

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

Consider the following relationships between the variables x and y:

$x^2+4=-6y$

$y^2+9=-4x$

Which answer is correct?

_{^{$(x+2)^2+(y+3)^2=0$}}

Solve the following system of equations:

$\begin{cases} \sqrt{x}+\sqrt{y}=\sqrt{\sqrt{61}+6} \\ xy=9 \end{cases}$

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

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

or

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

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