Solve for X: Triangle with Sides (x+1), (x+2), and x

In order to find the unknown in the problem, let's first recall the Pythagorean theorem which states that the sum of squares of the legs in a right triangle (the sides containing the right angle) equals the square of the hypotenuse (the side opposite to the right angle),

In other words, mathematically,

in a right triangle with legs of length: $a,b$ and hypotenuse of length:

$c$it is always true that:

$a^2+b^2=c^2$

Let's return then to the triangle given in the problem, from the triangle drawing we notice that the legs' lengths are:

$x,\hspace{2pt}x+1$

and the hypotenuse length is:

$x+2$

Therefore, according to the Pythagorean theorem we have:

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

Let's continue and solve the resulting equation, we'll start by simplifying the expressions on both sides,

For this we'll recall the square of binomial formula:

$(z\pm w)^2=z^2\pm2zw+w^2$

Let's apply this formula to the equation we got, first let's expand the parentheses, then combine like terms:

$x^2+\textcolor{red}{(x+1)^2}=\textcolor{blue}{(x+2)^2}\\ \downarrow\\ x^2+\textcolor{red}{x^2+2\cdot x\cdot1+1^2}=\textcolor{blue}{x^2+2\cdot x\cdot2+2^2}\\ x^2+x^2+2x+1=x^2+4x+4\\ x^2-2x-3=0$

We've therefore obtained a quadratic equation, we identify that the coefficient of the quadratic term is 1, so we can (try to) solve it using the quick factoring method,

Let's look for a pair of numbers whose product equals the constant term on the left side, and whose sum equals the coefficient of the first degree term meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=-3\\ m+n=-2$

From the first requirement above, namely the product, we can deduce according to the rules of sign multiplication that the two numbers have opposite signs, and now we'll remember that the only possible pair of factors of the (prime) number 3 are 3 and 1, fulfilling the second requirement mentioned, together with the fact that the numbers we're looking for have opposite signs will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=-3\\ n=1 \end{cases}$

Therefore we can factor the expression on the left side of the equation to:

$x^2-2x-3=0 \\ \downarrow\\ (x-3)(x+1)=0$

Where we used the pair of numbers we found earlier in this factorization,

Let's continue and consider the fact that on the left side of the last equation we have a product of algebraic expressions and on the right side we have 0, therefore, since the only way to get a product of 0 is to multiply by 0, at least one of the expressions in the product on the left side must equal zero,

Meaning:

$x-3=0\\ \boxed{x=3}$

Or:

$x+1=0\\ \boxed{x=-1}$

However, from the domain of definition for x specified in the problem:

x>1

We can eliminate the solution: $x=-1$Therefore the only solution to the unknown in the problem that satisfies the given condition is:

$\boxed{x=3}$

Therefore the correct answer is answer A.