Solve for x in the Triangle with Sides x, x+1, and x+2

To find $x$ in the given triangle, let's apply the Pythagorean Theorem. The squared lengths of the triangle's legs and hypotenuse are related by this equation:

First, expand each term:

• $(x+1)^2 = x^2 + 2x + 1$
• $x^2 = x^2$
• $(x+2)^2 = x^2 + 4x + 4$

Plug these into the Pythagorean Theorem equation:

Combine like terms:

Rearrange the equation to isolate terms on one side:

Simplify to get a quadratic equation:

Now, solve for $x$ using factoring. Look for two numbers that multiply to $-3$ and add to $-2$. These numbers are $-3$ and $1$:

Set each factor equal to zero:

• $x - 3 = 0 \Rightarrow x = 3$
• $x + 1 = 0 \Rightarrow x = -1$

Given the condition $x > 0$, the valid solution is:

$x = 3$