Solve for X: Finding Triangle Area When A = x²/2

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's drawing we notice that the lengths of its legs are:

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

and the length of the hypotenuse 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 shortened multiplication formula for squaring a binomial:

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

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

$x^2+\textcolor{red}{(x-1)^2}=\textcolor{blue}{(x+2)^2}\\ \downarrow\\ x^2+\textcolor{red}{x^2-2\cdot x\cdot1+1^2}=\textcolor{blue}{x^2+2\cdot x\cdot2+2^2}\\ x^2+x^2-2x+1=x^2+4x+4\\ x^2-6x-3=0$

(We'll stop here for now, continuation after data modification and correction)

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

We'll look for a pair of numbers whose product is the free term on the left side, and whose sum is 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 multiplication, we can deduce according to the rules of sign multiplication that the two numbers have different 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, along with the fact that the numbers we're looking for have different 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'll 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,

We'll continue and consider the fact that on the left side of the equation we got in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get a result of 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication 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 is:

$\boxed{x=3}$

Therefore the correct answer is answer A.