Solve the Triangle Equation: x with Constraint x > 0

To solve this problem, we'll use the Pythagorean Theorem to establish a relationship between the sides of the right triangle:

Given:

• One side $a = x$
• Another side $b = x + 7$
• The hypotenuse $c = 13$

According to the Pythagorean Theorem:

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

Substitute the given values:

$x^2 + (x + 7)^2 = 13^2$

Expand and simplify:\

$x^2 + (x^2 + 14x + 49) = 169$

$2x^2 + 14x + 49 = 169$

Subtract 169 from both sides to set the equation to 0:

$2x^2 + 14x + 49 - 169 = 0$

$2x^2 + 14x - 120 = 0$

Divide the entire equation by 2 to simplify:

$x^2 + 7x - 60 = 0$

We now have a quadratic equation that can be factored as:

$(x + 12)(x - 5) = 0$

Set each factor equal to 0 and solve for $x$:

• $x + 12 = 0$ gives $x = -12$
• $x - 5 = 0$ gives $x = 5$

Since $x > 0$, we have $x = 5$.

Therefore, the solution to the problem is $x = 5$.