Answer:
[tex]1 + \sqrt{14}[/tex]
Step-by-step explanation:
Using pythagorean theorem, x and x+6 are the legs, and x+7 is the hypotenuse.
You can express this as an equation.
[tex]x^2 + (x+6)^2 = (x+7)^2[/tex]
Expanding, this is [tex]x^2+x^2+12x+36 = x^2+14x+49[/tex]
Grouping up, this is [tex]2x^2+12x+36 = x^2+14x+49[/tex]
Rearranging, this is [tex]x^2-2x-13 = 0[/tex]
Now, we can use the quadratic formula to solve.
[tex]\frac{-(-2) \pm \sqrt{(-2)^2-4\cdot1\cdot(-13}}{2\cdot1}[/tex] = [tex]-1 \pm \sqrt{14}[/tex]. Since we need x to be positive (due to it being a side length), our answer is [tex]1 + \sqrt{14}[/tex], because it is the positive option
