Question 475477
If you know that 10, 24, 26 can be the sides of a right triangle, we've already found a solution. However, quadratics always have two solutions so we have to expand or show that x = 10 is the only solution.


Expanding, we get


*[tex \LARGE x^2 + (4x^2 + 16x + 16) = 676]


*[tex \LARGE 5x^2 + 16x - 660 = 0]


We know x = 10 works, so factor this out:


*[tex \LARGE 5(x-10)(x-r) = 0] where r is the other root. The constant term in our quadratic is -660, so 50r = -660, r = -66/5.


*[tex \LARGE 5x^2 + 16x - 660 = 5(x-10)(x + \frac{66}{5})]


The solutions are {10, -66/5}.