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:

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

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:

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

Combine like terms:

$2x^2 + 2x + 1 = x^2 + 4x + 4$

Rearrange the equation to isolate terms on one side:

$2x^2 + 2x + 1 - x^2 - 4x - 4 = 0$

Simplify to get a quadratic equation:

$x^2 - 2x - 3 = 0$

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

$(x - 3)(x + 1) = 0$

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:

$x = 3$