In #15 you need to apply the Pythagorean Theorem.
If we're dealing with a right triangle, then the sum of the squares of the 2 shorter sides of the triangle is equal to the square of the hypotenuse:
a^2 + b^2 = c^2
In your particular problem, a = x+6, b= x+3, and c = x+9.
Then (x+6)^2 + (x+3)^2 = (x+9)^2
so: x^2 + 12x + 36 + x^2 + 6x + 9 = x^2 + 18x + 81
Cancel x^2 from both sides. Then you have
12x + 36 + x^2 + 6x + 9 = 18x + 81
Now group all the x terms together on the left side:
12x + x^2 + 6x - 18x + 36 + 9 - 81 = 0
Combine all the x terms. Combine all the constants.
x^2 - 36 = 0 Factor this: (x-6)(x+6) = 0
Then x=6 and x=-6. We must check whether one or both of these results is correct before calling it a "solution."
Example: (-6+6)^2 + (-6+3)^2 = (-6+9)^2 True or false?
0 + 9 = 9 True! so x = -6 is a solution.
You MUST also check to determine whether or not 6 is also a solution.
