Some primitive triples are ... (3, 4, 5) (5, 12, 13) (7, 24, 25) (9, 40, 41) One interesting characteristic of these is that the sum of the last two numbers is the square of the first number.
Any multiple of these will be a Pythagorean triple.
Now consider your list. a) (10, 24, 26) = 2×(5, 12, 13) . . . IS a Pythagorean Triple b) 2×(7, 24, 25) = (14, 48, 50), so (14, 48, 49) is NOT a Pythagorean Triple c) 3×(3, 4, 5) = (9, 12, 15), so (9, 12, 16) is NOT a Pythagorean Triple d) (9, 40, 41) . . . IS a Pythagorean Triple e) 5×(3, 4, 5) = (15, 20, 25) . . . IS a Pythagorean Triple
The sets of side lengths that are Pythagorean Triples are ... (10, 24, 26) (9, 40, 41) (15, 20, 25)
<span>Here let the quadratic equation be ax^2 + bx + c
We know that a=5 from the question.
Since the roots are 6 and 2, the quadratic equation would take the form of a product like (a1x-b1)(a2x-b2).
However, let's assume that a2=1 and b2=6,
Since a=5, a1=5, then 5x-b1=5(x-2). Solving this shows that b1=10
So, the equation is (5x-10)(x-6)</span>
