Square of sum: Using multiple rules

Let's examine the given equation:

$(x+3)^2=(x-3)^2$First, let's simplify the equation, for this we'll use the perfect square formula for a binomial squared:

$(a\pm b)^2=a^2\pm2ab+b^2$,

We'll start by opening the parentheses on both sides simultaneously using the perfect square formula mentioned, then we'll move terms and combine like terms, and in the final step we'll solve the simplified equation we get:

$(x+3)^2=(x-3)^2 \\ \downarrow\\ x^2+2\cdot x\cdot3+3^2= x^2-2\cdot x\cdot3+3^2 \\ x^2+6x+9= x^2-6x+9 \\ x^2+6x+9- x^2+6x-9 =0\\ 12x=0\hspace{6pt}\text{/}:12\\ \boxed{x=0}$Therefore, the correct answer is answer A.

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:

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

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